Question

In the figure , if QR =24 m, NR = 4 m and MN = 6 m. then find the length of PQ.. ​

Answer 1

Given:

In the figure, if QR =24 m, NR = 4 m and MN = 6 m. then find the length of PQ.

To find:

The length of PQ.​

Solution:

In Δ RNM and Δ RQP, we get

∠MRN = ∠PRQ . . . [common angle]

∠RNM = ∠RQP = 90° . . . [from the figure]

∴ Δ RNM ~  Δ RQP . . . [by AA similarity]

We know that →

The corresponding sides of the similar triangles are proportional to each other.

So, based on the above theorem, for the two similar triangles ΔRNM and Δ RQP, we get

$\frac{NR}{QR} = \frac{MN}{PQ}$

on substituting the given values of QR =24 m, NR = 4 m and MN = 6 m, we get

$\implies \frac{4}{24} = \frac{6}{PQ}$

$\implies \frac{1}{6} = \frac{6}{PQ}$

$\implies PQ = 6\times 6$

$\implies \bold{PQ = 36\:m}$

Thus, the length of PQ is → 36 m.

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Also View:

If AD and PM are medians of triangles ABC and PQR, respectively where ΔABC ∼ ΔPQR prove that AB/PQ = AD/PM

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in the triangle ABC similar to triangle ade. if ad:db=2:3 and de=5cm. 1) find bc. 2) if x be the length of the perpendicular from a to be, find the length of the perpendicular from a to bc in terms of x

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ABC and ADE are two similar triangles. If AD: DB=2:3 and DE = 5 cm, calculate length of BC. Find the altitude of ABC if the altitude of ADE is h.

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Answer 2

Answer:

Given: PQ=10cm, PR=24cm

Let QR be x cm.

In right angled triangle QPR,

(Hypotenuse)

2

=(Base)

2

+(Perpendicular)

2

[By Pythagoras theorem]

⇒(QR)

2

=(PQ)

2

+(PR)

2

⇒x

2

=(10)

2

+(24)

2

⇒x

2

=100+576=676

⇒x=

676

=26cm

Thus, the length of QR is 26cm.

solution