- ABC- PQR- M is the mid point of BC and N is the mid point of QR- If the area of - PQR - 100 sq-cm- and the area of - 144 sq-cm- If AM - 4cm- then PN is -

The correct option is D 3.3 cmSince the triangles are similar, (AM) ^ 2 (PN) ^ 2 = area of Δ ABC area of Δ PQR ⟹ (4) ^ 2 (PN) ^ 2 = ...

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Byju's Answer

Standard VIII

Mathematics

Dilations

Δ ABC∼Δ PQR. ...

Question

$Δ A B C \sim Δ P Q R$ . M is the mid point of BC and N is the mid point of QR. If the area of $Δ P Q R$ = 100 sq.cm. and the area of = 144 sq.cm. If AM = 4cm, then PN is :

3.3 cm

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Solution

The correct option is D 3.3 cm
Since the triangles are similar,
$\frac{{(A M)}^{2}}{{(P N)}^{2}} = \frac{a r e a o f Δ A B C}{a r e a o f Δ P Q R} ⟹ \frac{{(4)}^{2}}{{(P N)}^{2}} = \frac{144}{100} ⟹ \frac{16}{{(P N)}^{2}} = \frac{144}{100} ⟹ \frac{1}{{(P N)}^{2}} = \frac{9}{100} ⟹ P N = \frac{10}{3} = 3.3$

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Similar questions

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Δ A B C \sim Δ P Q R

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ΔABC∼ΔPQR. M is the mid point of BC and N is the mid point of QR. If the area of ΔPQR = 100 sq.cm. and the area of = 144 sq.cm. If AM = 4cm, then PN is :

The correct option is D 3.3 cmSince the triangles are similar,(AM)2(PN)2=areaofΔABCareaofΔPQR⟹(4)2(PN)2=144100⟹16(PN)2=144100⟹1(PN)2=9100⟹PN=103=3.3

$Δ A B C \sim Δ P Q R$ . M is the mid point of BC and N is the mid point of QR. If the area of $Δ P Q R$ = 100 sq.cm. and the area of = 144 sq.cm. If AM = 4cm, then PN is :