Area Class 8 Short Question Answer

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Maths Class 8 Maths 103 views Jun 17, 2026 Reviewed & updated Sep 17, 2026

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Area Class 8 Short Question Answer

Area Class 8 Short Question Answer

Question 1.

Consider the obtuse angled triangle ABC of base 6 cm.

Its height AD which is perpendicular from the vertex A is outside the triangle.

Can you find the area of the triangle ?

Solution:

Yes, its area can be found out.

Area of ∆ABC = [latex]\frac{1}{2}[/latex] × Base × Height

= [latex]\frac{1}{2}[/latex] × BC × AD ²

= ([latex]\frac{1}{2}[/latex] × 6 × 4) cm² = 12cm².

Question 2.

Find the area of the following paral-lelogram :

Solution:

Area of the parallelogram = Base × Height

= (2.5 × 3.5) cm².

= 8.75 cm².

Question 3.

DL and BM are the heights on sides AB and AD respectively of parallelogram ABCD. If the area of the parallelogram is 1470 cm², AB = 35 cm and AD = 49 cm, find the lengths of BM and DL.

Solution:

We have :

Area of parallelogram ABCD

= 1470 cm²

AB = 35 cm and AD = 49 cm

Area of parallelogram ABCD

= AD × BM

So, 1470 = 49 × BM

or BM = [latex]\frac{1470}{49}[/latex] cm

= 30 cm.

Again,

Area of parallelogram ABCD

= AB × DL

So, 1470 = 35 × DL

or DL = [latex]\frac{1470}{35}[/latex] cm

= 42 cm.

Question 4.

∆ABC is right angled at A. AD is perpendicular to BC. If AB = 5 cm, BC = 13 cm and AC = 12 cm. Find the area of ∆ABC. Also, find the length of AD.

Solution:

Since ∆ ABC is right ∠d at A, therefore

Area of ∆ABC = [latex]\frac{1}{2}[/latex] × AB × AC

= ([latex]\frac{1}{2}[/latex] × 5 × 12) cm²

= 30 cm².

Again, area of ∆ABC = [latex]\frac{1}{2}[/latex] × BC × AD

So, 30 = [latex]\frac{1}{2}[/latex] × 13 × AD

or AD = [latex]\frac{2 \times 30}{13}[/latex] cm

= [latex]\frac{60}{30}[/latex] cm.

Question 5.

∆ABC is isosceles with AB = AC = 7.5 cm and BC = 9 cm. The height AD from A to BC is 6 cm. Find the area of ∆ABC. What will be the height from C to AB i.e., CE?