Ex.11.2 Q4 Perimeter and Area - NCERT Maths Class 7

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Question

Find the missing values:

S. No. Base Height Area of Triangle
a. \(15 \,\rm cm\)   \(87 \,\rm cm^2\)
b.   \(31.4 \,\rm mm\) \(1256\,\rm mm^2\)
c. \(22 \,\rm cm\)   \(170.5 \,\rm cm^2\)

 Video Solution
Perimeter And Area
Ex 11.2 | Question 4

Text Solution

What is known?

Area of triangle.

What is unknown?

Base and height of the triangle.

Reasoning:

In this question, area of the triangles are given. In some parts of the question base of the triangle and in some other parts height of the triangle is given. Put the values of the given part directly into the formula of area of triangle and find the unknown value.

Steps:

(a) Given,

Area of the triangle \(= 87 \,\rm cm^2\)

Base \(= 15 \,\rm cm\)

Height \(=\,?\)

We know that,

\(\begin{align}&{\text{Area of triangle}} \\ &= \frac{1}{2} \times {\rm{Base}} \times {\rm{Height}} \end{align} \)

\(\begin{align}87 &= \frac{1}{2} \times 15 \times {\rm{Height}}\\\\{\rm{Height}} &= \frac{{2 \times 87}}{{15}}\\\\{\rm{Height}} &= \frac{{174}}{{15}}\\\\{\rm{Height}} &= 11.6\;\rm{cm}\end{align}\)

(b) Given,

Area of the triangle \(= 1256\,\rm mm^2\)

Height\(= 31.4 \,\rm mm\)

Base \(=\,?\)

We know that,

\(\begin{align}&{\text{Area of triangle}} \\ &= \frac{1}{2} \times {\rm{Base}} \times {\rm{Height}} \end{align} \)

\(\begin{align}1256 &= \frac{1}{2} \times {\rm{Base}} \times {\rm{31}}{\rm{.4}}\\\\{\rm{Base}} &= \frac{{2 \times 1256}}{{31.4}}\;\\\\{\rm{Base}} &= \frac{{2512}}{{31.4}}\\\\{\rm{Base}} &= 80\; \rm mm\\\end{align}\)

(c) Given,

Area of the triangle \(= 170.5 \,\rm cm^2\)

Base \(= 22 \,\rm cm\)

Height \(=\,?\)

We know that,

\(\begin{align}&{\text{Area of triangle}} \\ &= \frac{1}{2} \times {\rm{Base}} \times {\rm{Height}} \end{align} \)

\(\begin{align}170.5 &= \frac{1}{2} \times {\rm{22}} \times {\rm{Height}}\\{\rm{Height}} &= \frac{{2 \times 170.5}}{{22}}\\{\rm{Height}} &= \frac{{341}}{{22}}\\{\rm{Height}} &= 15.5\; \rm cm\end{align}\)

  
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