Similar Triangles
Friends, maine aapko apne previous article mein similar triangles ke baare mein short mein bataya tha. Friends, maine uss article mein similar triangles ke liye different criteria ke baare mein bhi bataya tha. Inn criteria ka use karke hum ye bata sakte hai ki diye gaye do(two) ya do(two) se jyada triangles similar hain ya nahi.
Friends aaj me apne iss article mein aapko similar triangles ke baare mein detail mein bataoonga aur hum iss article mein similar triangles ke areas ke baare mein bhi discuss karenge.
Friends, mera naam hai Dheeraj Sahni aur aap log iss waqt hain meri website www.mathshindi.com par. Agar aap similar triangles ke baare mein jaanna chahte hain to bane rahiye mere saath iss article ke last tak...
Inn pictures ko dekhkar aap ye turant (at once) bata sakte hain ki ye teeno(three) pictures ek hi monument- Taj Mahal ki hai lekin inn pictures ke size alag-alag(different) hain. Taj Mahal ki ye teeno(three) pictures similar kahe jaayenge. In fact koi bhi do(two) ya do(two) se jyada cheezein jo shape mein me to same ho (lekin ye zaroori nahi ki size me bhi same ho) similar cheezein kehte hain.
Isi tarah se koi bhi do(two) ya do(two) se jyada figures (chahe wo figures triangles ho, chahe wo figures rectangle ho, chahe wo figures circle ho, chahe wo figures pentagon ho etc.) similar hote hain jab unke shape same ho but ye zaroori nahi ki unke size bhi same ho.
Koi bhi do(two) aise figures jisme number of sides same ho, hi similar ho sakte hain. Jinn figures mein number of sides same naa ho, wo figures kabhi bhi similar nahi ho sakte hain. Example ke liye neeche diye gaye inn do(two) diagrams ko dekhiye-
(i) (ii)
Pehla figure [figure (i)] ek circle hai jisme ek bhi side nahi hoti hai yaani ki circle mein 0 side hoti hai. Aur doosra figure [figure (ii)] ek triangle hai jisme 3 sides hoti hain. So, yeh dono figures (circle and triangle) kabhi bhi similar figures nahi ho sakte hain.
To chaliye dekhte hain similar figures ki definition-
“Two polygons of the same number of sides are similar, if
(i) their corresponding angles are equal and
(ii) their corresponding sides are in the same ratio (or proportion)”.
Matlab koi bhi diye gaye do(two) polygons, jinme number of sides same ho, similar hote hain agar,
(i) unke corresponding angles equal ho aur
(ii) unke corresponding sides same ratio (proportion) mein ho.
Example ke liye neeche diye gaye iss diagram ko dekhiye-
Iss diagram me dono quadrilateral (jo ki ek polygon hai) ke corresponding angles same hai, matlab ∠A = ∠P = 105°,∠B = ∠Q = 100°, ∠C = ∠R = 70°, ∠D = ∠S = 85° aur corresponding sides bhi same ratio (proportion) mein hai, matlab
AB = 1.5 = 1, BC = 2.5 = 1, CD = 2.4 = 1,
PQ 3.0 2 QR 5.0 2 RS 4.8 2
DA = 2.1 = 1 .
SP 4.2 2
[Note: Corresponding angles ka matlab hota hai dono triangles mein wo angles jo same jagah/place par aate hain. Jaise ki iss diagram ko dekhiye-
Iss diagram me triangle ΔABC mein angle ∠A ko lijiye. Ab triangle ΔDEF ko dekhiye ΔDEF mein ∠A ki jagah/place par ∠D hai. So, ∠A aur ∠D ek doosre ke corresponding angle hue. And ∠B aur ∠E, ∠C aur ∠F ek doosre ke corresponding angle hue. Isi tarah se corresponding sides bhi hoti hai AB aur DE, AC aur DF, BC aur EF ek doosre ke corresponding sides hain].
Iss article mein hum saare polygons ki baat nahi karenge. Hum iss article mein baat karenge sirf triangles ki. To chaliye dekhte hain ki do(two) triangles kab similar hote hain aur triangles ke similarity ki definition kya hai–
“Two triangles are similar, if
(i) their corresponding angles are equal and
(ii) their corresponding sides are in the same ratio (proportion)”.
Matlab koi bhi diye gaye do(two) triangles similar hote hai agar,
(i) unke corresponding angles equal ho aur
(ii) unke corresponding sides same ratio (proportion) mein ho.
Example ke liye neeche diye gaye triangles ke inn do(two) figures ko dekhiye–
In dono triangles ke teeno corresponding angles equal hain matlab ∠L =∠S = 40°, ∠M = ∠T = 60°, ∠N = ∠U = 80°. Aur teeno corresponding sides ke ratio equal hain matlab
LM = 24 = 2 ,
ST 12 1
MN = 28 = 2 aur NL = 18 = 2 .
TU 14 1 US 9 1
Isliye ye dono triangles, ΔLMN aur ΔSTU similar triangles hue. Iss statement ko hum symbolically is tarah se likhte-
ΔLMN∼ΔSTU. Iska matlab ye hai ki triangle ΔLMN aur ΔSTU similar hain.
[Note: Corresponding sides ke same ratio ko scale factor ( ya representative fraction) kehte hain].
Triangles ki similarity check karne ke liye kuch methods/ways/criteria/theorems hain. Inn methods/ways/criteria/theorems ka use karke hum ye aasaani se bata sakte hain ki diye gaye do(two) triangles similar hain ya nahi. Ye chaar(four) methods/ways/criteria/theorems hain:
(i) AAA similarity criterion
(ii) AA similarity criterion
(iii) SSS similarity criterion
(iv) SAS similarity criterion
AAA similarity criterion me ye hota hai ki agar diye gaye do(two) triangles ke teeno corresponding angles equal hain to wo dono triangles similar hote hain.
AA similarity criterion mein ye hota hai ki agar diye gaye do(two) triangles ke koi se do(two) corresponding angles equal hai to wo dono triangles similar hote hain.
SSS similarity criterion me ye hota hai ki agar diye gaye do(two) triangles ke teeno corresponding sides ke ratio equal hai to dono triangles similar honge.
SAS similarity criterion me ye hota hai ki agar diye gaye do(two) triangles ke koi si bhi do(two) corresponding sides same ratio mein ho aur inn dono sides ke beech ka angle jisko included angle kehte hain bhi equal ho to wo dono triangles similar hote hain.
[Agar aap inn chaaro(four) criteria ke baare mein aur jyada jaanna chahte hain to aap hamare iss post ko padhiye »»»
Criteria for Similarity of Triangles].
Also Read:
Basic Proportionality Theorem (Thales Theorem)
To abhi aapne padha ki do(two) similar triangles mein, corresponding sides ke ratio same hote hain. Aap kya sochte hain? Kya similar triangles mein unke corresponding sides ke ratio aur unke areas ke ratio ke beech mein kuch relation hai? Iska jawaab aapko iss theorem ko padhkar milega:
Theorem
“The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.”
[Do(two) similar triangles ke areas ka ratio, unke corresponding sides ke ratio ke square ke equal hota hai].
Proof of the Theorem
Given: Do(two) triangles, ΔABC aur ΔPQR diye hue hain aur ΔABC∼ΔPQR.
To prove:
Construction: Altitude AM aur PN draw kar diye jisse ki triangles ka area nikaala jaa sake.
Proof: Now, ar(ΔABC) = (1/2) × BC × AM
and ar(ΔPQR) = (1/2) × QR × PN
Therefore, ar(ΔABC) = (1/2) × BC × AM
ar(ΔPQR) (1/2) × QR × PN
= BC × AM -------(1)
DB × PN
Now, in ΔABM and ΔPQN,
∠B = ∠Q (ABCPQR)
and ∠M= ∠N (each 90°)
So, ΔABM∼ΔPQN (AA similarity criterion)
Therefore, AM = AB --------(2)
PN PQ
Now, ΔABC∼ΔPQR (Given)
So, AB = BC = CA ---------(3)
PQ QR RP
Therefore, ar(ΔABC) = AB × AM [From (1) and (2)]
ar(ΔPQR) PQ PN
= AB × AB [From (1)]
PQ PQ
Now, using (3) we get,
Chaliye iss article mein hamne abhi jo kuch bhi padha usse related kuch solved examples dekh lete hain jisse ki iss article ke concepts aapko aur achche se samajh mein aa jaaye.
Ex.1. In the given figure, if PQ∥RS, prove that ΔPOQ∼ΔSOR.
Solution: PQ∥RS (Given)
So, ∠P = ∠S (Alternate angles)
and ∠Q = ∠R (Alternate angles) Now, ∠POQ = ∠SOR (Vertically opposite angles)
Therefore, ΔPOQ∼ΔSOR (AAA similarity criterion)
Ex.2. Observe the figure and then find P.
Solution: In ΔABC and ΔPQR,
AB = 3.8 = 1 ,
RQ 7.6 2
BC = 6 = 1 ,
QP 12 2
and CA = 3√3 = 1 .
PR 6√3 2
That means, AB = BC = CA
RQ QP PR
So, ΔABC∼ΔRQP (SSS similarity criterion)
Therefore, ∠C = ∠P (Corresponding angles of similar triangles)
But, ∠C = 180° – ∠A – ∠B (Angle sum property)
= 180° – 80° – 60°
= 180° – 140°
= 40°
So, ∠P = ∠C = 40°
Ex.3. In the given figure,
OA • OB = OC • OD. Show that ∠A = ∠C and ∠B = ∠D.
Solution: OA • OB = OC • OD (Given)
So, OA = OD --------(1)
OC OB
Now, ∠AOD = ∠COB (Vertically opposite angles) ---------(2)
So, from (1) and (2),
ΔAOD∼ΔCOB (SAS similarity criterion)
So, ∠A= ∠C and ∠D = ∠B (Corresponding angles of similar triangles)
Ex. 4. A girl of height 90 cm is walking away from the base of a lamp-post at a speed of 1.2 m/s. If the length is 3.6 m above the ground, find the length of her shadow after 4 seconds.
Solution: Maan lijiye ki AB lamp-post ko denote karta hai aur CD, 4 second chalne ke baad, girl ko denote karta hai (diagram dekhiye).
Diagram se, aap dekh sakte hain ki DE girl ki shadow hai. Maan lijiye ki DE, x metre hai.
Now, BD = 1.2 m × 4 = 4.8 m
Now, in ΔABE and ΔCDE,
∠B = ∠D [Each is 90°, kyuki lamp-post aur girl dono ground ke vertically khade(stand) hain].
∠E = ∠E (Same angle)
ΔABE∼ΔCDE (AA similarity criterion)
Therefore, BE = AB
DE CD
i.e., 4.8 + x = 3.6 (90 cm = 0.9 m)
x 0.9
i.e., 4.8 + x = 4x
i.e., 4.8 = 4x – x
i.e., 4.8 = 3x
i.e., 4.8 = x
3
i.e., 1.6 = x
So, the shadow of the girl walking for 4 seconds is 1.6 m long.
Ex. 5. Sides of two similar triangles are in the ratio 4:9. What ratio are the areas of these triangles in ?
Solution: One side of first triangle = 4
One side of second triangle 9
= 16
81
= 16:81
I hope ki mera ye article aapko pasand aaya hoga. Agar aapko mera ye article pasand aaya to aap hame comment ke maadhyam se bata sakte hain.
Aapne is article ko padhkar kya seekha?
CHECK YOUR KNOWLEDGE
Give answers:
Q.1- Is a triangle and a square can be similar figures?
Q.2- If the areas of two similar triangles are equal, prove that they are congruent.
Q.3- ΔABC and ΔBDE are two equilateral triangles such that D is the mid-point of BC. Find the ratio of the areas of triangles, ΔABC and ΔBDE.
Aap apne answers hame comment ke through bata sakte hain.
THANKS FOR READING THIS BLOG.
Friends aaj me apne iss article mein aapko similar triangles ke baare mein detail mein bataoonga aur hum iss article mein similar triangles ke areas ke baare mein bhi discuss karenge.
Friends, mera naam hai Dheeraj Sahni aur aap log iss waqt hain meri website www.mathshindi.com par. Agar aap similar triangles ke baare mein jaanna chahte hain to bane rahiye mere saath iss article ke last tak...
Let's begin...
Friends hum apni day-to-day life mein aisi cheezon se milte hain jo ki similar hoti hain. Example ke liye inn teen(three) pictures ko dekhiye jo ki neeche diye gaye hain:
Inn pictures ko dekhkar aap ye turant (at once) bata sakte hain ki ye teeno(three) pictures ek hi monument- Taj Mahal ki hai lekin inn pictures ke size alag-alag(different) hain. Taj Mahal ki ye teeno(three) pictures similar kahe jaayenge. In fact koi bhi do(two) ya do(two) se jyada cheezein jo shape mein me to same ho (lekin ye zaroori nahi ki size me bhi same ho) similar cheezein kehte hain.
Isi tarah se koi bhi do(two) ya do(two) se jyada figures (chahe wo figures triangles ho, chahe wo figures rectangle ho, chahe wo figures circle ho, chahe wo figures pentagon ho etc.) similar hote hain jab unke shape same ho but ye zaroori nahi ki unke size bhi same ho.
Koi bhi do(two) aise figures jisme number of sides same ho, hi similar ho sakte hain. Jinn figures mein number of sides same naa ho, wo figures kabhi bhi similar nahi ho sakte hain. Example ke liye neeche diye gaye inn do(two) diagrams ko dekhiye-
Pehla figure [figure (i)] ek circle hai jisme ek bhi side nahi hoti hai yaani ki circle mein 0 side hoti hai. Aur doosra figure [figure (ii)] ek triangle hai jisme 3 sides hoti hain. So, yeh dono figures (circle and triangle) kabhi bhi similar figures nahi ho sakte hain.
To chaliye dekhte hain similar figures ki definition-
“Two polygons of the same number of sides are similar, if
(i) their corresponding angles are equal and
(ii) their corresponding sides are in the same ratio (or proportion)”.
Matlab koi bhi diye gaye do(two) polygons, jinme number of sides same ho, similar hote hain agar,
(i) unke corresponding angles equal ho aur
(ii) unke corresponding sides same ratio (proportion) mein ho.
Example ke liye neeche diye gaye iss diagram ko dekhiye-
Iss diagram me dono quadrilateral (jo ki ek polygon hai) ke corresponding angles same hai, matlab ∠A = ∠P = 105°,∠B = ∠Q = 100°, ∠C = ∠R = 70°, ∠D = ∠S = 85° aur corresponding sides bhi same ratio (proportion) mein hai, matlab
AB = 1.5 = 1, BC = 2.5 = 1, CD = 2.4 = 1,
PQ 3.0 2 QR 5.0 2 RS 4.8 2
DA = 2.1 = 1 .
SP 4.2 2
[Note: Corresponding angles ka matlab hota hai dono triangles mein wo angles jo same jagah/place par aate hain. Jaise ki iss diagram ko dekhiye-
Iss diagram me triangle ΔABC mein angle ∠A ko lijiye. Ab triangle ΔDEF ko dekhiye ΔDEF mein ∠A ki jagah/place par ∠D hai. So, ∠A aur ∠D ek doosre ke corresponding angle hue. And ∠B aur ∠E, ∠C aur ∠F ek doosre ke corresponding angle hue. Isi tarah se corresponding sides bhi hoti hai AB aur DE, AC aur DF, BC aur EF ek doosre ke corresponding sides hain].
Iss article mein hum saare polygons ki baat nahi karenge. Hum iss article mein baat karenge sirf triangles ki. To chaliye dekhte hain ki do(two) triangles kab similar hote hain aur triangles ke similarity ki definition kya hai–
“Two triangles are similar, if
(i) their corresponding angles are equal and
(ii) their corresponding sides are in the same ratio (proportion)”.
Matlab koi bhi diye gaye do(two) triangles similar hote hai agar,
(i) unke corresponding angles equal ho aur
(ii) unke corresponding sides same ratio (proportion) mein ho.
Example ke liye neeche diye gaye triangles ke inn do(two) figures ko dekhiye–
In dono triangles ke teeno corresponding angles equal hain matlab ∠L =∠S = 40°, ∠M = ∠T = 60°, ∠N = ∠U = 80°. Aur teeno corresponding sides ke ratio equal hain matlab
LM = 24 = 2 ,
ST 12 1
MN = 28 = 2 aur NL = 18 = 2 .
TU 14 1 US 9 1
Isliye ye dono triangles, ΔLMN aur ΔSTU similar triangles hue. Iss statement ko hum symbolically is tarah se likhte-
ΔLMN∼ΔSTU. Iska matlab ye hai ki triangle ΔLMN aur ΔSTU similar hain.
[Note: Corresponding sides ke same ratio ko scale factor ( ya representative fraction) kehte hain].
Triangles ki similarity check karne ke liye kuch methods/ways/criteria/theorems hain. Inn methods/ways/criteria/theorems ka use karke hum ye aasaani se bata sakte hain ki diye gaye do(two) triangles similar hain ya nahi. Ye chaar(four) methods/ways/criteria/theorems hain:
(i) AAA similarity criterion
(ii) AA similarity criterion
(iii) SSS similarity criterion
(iv) SAS similarity criterion
AAA similarity criterion me ye hota hai ki agar diye gaye do(two) triangles ke teeno corresponding angles equal hain to wo dono triangles similar hote hain.
AA similarity criterion mein ye hota hai ki agar diye gaye do(two) triangles ke koi se do(two) corresponding angles equal hai to wo dono triangles similar hote hain.
SSS similarity criterion me ye hota hai ki agar diye gaye do(two) triangles ke teeno corresponding sides ke ratio equal hai to dono triangles similar honge.
SAS similarity criterion me ye hota hai ki agar diye gaye do(two) triangles ke koi si bhi do(two) corresponding sides same ratio mein ho aur inn dono sides ke beech ka angle jisko included angle kehte hain bhi equal ho to wo dono triangles similar hote hain.
[Agar aap inn chaaro(four) criteria ke baare mein aur jyada jaanna chahte hain to aap hamare iss post ko padhiye »»»
Criteria for Similarity of Triangles].
Also Read:
Basic Proportionality Theorem (Thales Theorem)
To abhi aapne padha ki do(two) similar triangles mein, corresponding sides ke ratio same hote hain. Aap kya sochte hain? Kya similar triangles mein unke corresponding sides ke ratio aur unke areas ke ratio ke beech mein kuch relation hai? Iska jawaab aapko iss theorem ko padhkar milega:
Theorem
“The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.”
[Do(two) similar triangles ke areas ka ratio, unke corresponding sides ke ratio ke square ke equal hota hai].
Proof of the Theorem
Given: Do(two) triangles, ΔABC aur ΔPQR diye hue hain aur ΔABC∼ΔPQR.
To prove:
Construction: Altitude AM aur PN draw kar diye jisse ki triangles ka area nikaala jaa sake.
Proof: Now, ar(ΔABC) = (1/2) × BC × AM
and ar(ΔPQR) = (1/2) × QR × PN
Therefore, ar(ΔABC) = (1/2) × BC × AM
ar(ΔPQR) (1/2) × QR × PN
= BC × AM -------(1)
DB × PN
Now, in ΔABM and ΔPQN,
∠B = ∠Q (ABCPQR)
and ∠M= ∠N (each 90°)
So, ΔABM∼ΔPQN (AA similarity criterion)
Therefore, AM = AB --------(2)
PN PQ
Now, ΔABC∼ΔPQR (Given)
So, AB = BC = CA ---------(3)
PQ QR RP
Therefore, ar(ΔABC) = AB × AM [From (1) and (2)]
ar(ΔPQR) PQ PN
= AB × AB [From (1)]
PQ PQ
Now, using (3) we get,
Chaliye iss article mein hamne abhi jo kuch bhi padha usse related kuch solved examples dekh lete hain jisse ki iss article ke concepts aapko aur achche se samajh mein aa jaaye.
Ex.1. In the given figure, if PQ∥RS, prove that ΔPOQ∼ΔSOR.
Solution: PQ∥RS (Given)
So, ∠P = ∠S (Alternate angles)
and ∠Q = ∠R (Alternate angles) Now, ∠POQ = ∠SOR (Vertically opposite angles)
Therefore, ΔPOQ∼ΔSOR (AAA similarity criterion)
Ex.2. Observe the figure and then find P.
Solution: In ΔABC and ΔPQR,
AB = 3.8 = 1 ,
RQ 7.6 2
BC = 6 = 1 ,
QP 12 2
and CA = 3√3 = 1 .
PR 6√3 2
That means, AB = BC = CA
RQ QP PR
So, ΔABC∼ΔRQP (SSS similarity criterion)
Therefore, ∠C = ∠P (Corresponding angles of similar triangles)
But, ∠C = 180° – ∠A – ∠B (Angle sum property)
= 180° – 80° – 60°
= 180° – 140°
= 40°
So, ∠P = ∠C = 40°
Ex.3. In the given figure,
OA • OB = OC • OD. Show that ∠A = ∠C and ∠B = ∠D.
Solution: OA • OB = OC • OD (Given)
So, OA = OD --------(1)
OC OB
Now, ∠AOD = ∠COB (Vertically opposite angles) ---------(2)
So, from (1) and (2),
ΔAOD∼ΔCOB (SAS similarity criterion)
So, ∠A= ∠C and ∠D = ∠B (Corresponding angles of similar triangles)
Ex. 4. A girl of height 90 cm is walking away from the base of a lamp-post at a speed of 1.2 m/s. If the length is 3.6 m above the ground, find the length of her shadow after 4 seconds.
Solution: Maan lijiye ki AB lamp-post ko denote karta hai aur CD, 4 second chalne ke baad, girl ko denote karta hai (diagram dekhiye).
Diagram se, aap dekh sakte hain ki DE girl ki shadow hai. Maan lijiye ki DE, x metre hai.
Now, BD = 1.2 m × 4 = 4.8 m
Now, in ΔABE and ΔCDE,
∠B = ∠D [Each is 90°, kyuki lamp-post aur girl dono ground ke vertically khade(stand) hain].
∠E = ∠E (Same angle)
ΔABE∼ΔCDE (AA similarity criterion)
Therefore, BE = AB
DE CD
i.e., 4.8 + x = 3.6 (90 cm = 0.9 m)
x 0.9
i.e., 4.8 + x = 4x
i.e., 4.8 = 4x – x
i.e., 4.8 = 3x
i.e., 4.8 = x
3
i.e., 1.6 = x
So, the shadow of the girl walking for 4 seconds is 1.6 m long.
Ex. 5. Sides of two similar triangles are in the ratio 4:9. What ratio are the areas of these triangles in ?
Solution: One side of first triangle = 4
One side of second triangle 9
= 16
81
= 16:81
I hope ki mera ye article aapko pasand aaya hoga. Agar aapko mera ye article pasand aaya to aap hame comment ke maadhyam se bata sakte hain.
Aapne is article ko padhkar kya seekha?
CHECK YOUR KNOWLEDGE
Give answers:
Q.1- Is a triangle and a square can be similar figures?
Q.2- If the areas of two similar triangles are equal, prove that they are congruent.
Q.3- ΔABC and ΔBDE are two equilateral triangles such that D is the mid-point of BC. Find the ratio of the areas of triangles, ΔABC and ΔBDE.
Aap apne answers hame comment ke through bata sakte hain.
THANKS FOR READING THIS BLOG.
Comments