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PS = 3 cm
and, SR = 4 cm
So, PR = 4 + 3 = 7 cm
Now, ST║RQ
so, triangles PST and PRQ are similar [∵Alternate angles are equal]
Hence, ar(PST)/ar(PRQ) = PS²/PR² = 3²/7² = 9/49 = 9:49
and, SR = 4 cm
So, PR = 4 + 3 = 7 cm
Now, ST║RQ
so, triangles PST and PRQ are similar [∵Alternate angles are equal]
Hence, ar(PST)/ar(PRQ) = PS²/PR² = 3²/7² = 9/49 = 9:49
Answered by
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Given:
PS = 3 cm
SR = 4 cm
PR = PS + SR = 4 + 3 = 7 cm
ST║QR
To Find:
Ratio of area of ΔPST and ΔPRQ
Solution:
To find the ratio of PS and PR
= PS/SR ⇒ 3/7
( Note:- We don't want to use BPT here as it is not required )
as ST║QR It gives us that "ΔPST similar to ΔPQR"
Area of ΔPST/Area of ΔPQR = PS²/SR²
(By Theorem - Ratio of Area of similar triangles equals to their square of corresponding sides)
Area of ΔPST/Area of ΔPQR = 3²/7²
= 9/49
Hence their ratio of areas = 9:49
PS = 3 cm
SR = 4 cm
PR = PS + SR = 4 + 3 = 7 cm
ST║QR
To Find:
Ratio of area of ΔPST and ΔPRQ
Solution:
To find the ratio of PS and PR
= PS/SR ⇒ 3/7
( Note:- We don't want to use BPT here as it is not required )
as ST║QR It gives us that "ΔPST similar to ΔPQR"
Area of ΔPST/Area of ΔPQR = PS²/SR²
(By Theorem - Ratio of Area of similar triangles equals to their square of corresponding sides)
Area of ΔPST/Area of ΔPQR = 3²/7²
= 9/49
Hence their ratio of areas = 9:49
☺ Hope this Helps ☺
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