Question

Please solve question no. 8 and 10...
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Answer 1

Sol8.

Given:
AB = 6cm
BC = 8cm

To Find : BN

Step By step Sol.

As we can see that AB is the perpendicular and BC is Base. So, with Pythagoras Theorum we can find the Hypotenuse AC.

AB^2 + BC^2 = AC^2
6^2 + 8^2 = AC^2
36 + 64 = 100 ^2
AC = √100 = 10cm

Now, we should find the area of ΔABC by taking AB as height and BC as base.

So, Area of triangle =
$\frac{1}{2} \times base \times length$
$= \frac{1}{2} \times 6 \times 8$
$= 24 {cm}^{2}$
Now, we know that area of this ΔABC is 24 cm^2. Then, we can take AC as base and BN as the corresponding height. We can apply the area of triangle again and we will get the lenght of BN.

$\frac{1}{2} \times 10 \times bn = 24 {cm}^{2}$

$bn = \frac{24 \times 2}{10}$
$bn = \frac{48}{10}$
So, BN = 4.8


Sol 10.

Length = 10cm

Breadth = 18cm

Area of Rectangle = Lenght * Breadth
= 10 * 18 = 180cm^2

Now, we can see clearly that ΔBEC is a right-hand angled Δ as /_E is right angle. So, we can find the area of this right By taking EB as base and EC as height but first we need to find EC by Pythagoras Theorum.

We know that
EC is Perpendicular which is 8cm
EB is Base
CB is Hypotenuse which is 10cm

So, EC^2 + EB^2 = CB^2
8^2 + EB^2 = 10^2
64 + EB^2 = 100
EB = √100-64
EB = √36 = 6cm

Now, We can apply Area of

$\frac{1}{2} \times base \times height$
$\frac{1}{2} \times 8 \times 6$
$24 {cm}^{2}$
So, we can subtract the area of this right to get the area of the remaining figure.

180 - 24 = 156 cm^2