Question

Find the value of 'x' in the given figure. All measurements are given in centimetres.




Irrelevant answers will be reported.​

Answer 1

Answer:

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Step-by-step explanation:

LETS CONSUME BIGGER TRIANGLE AS ABC

AND SMALLER TRIANGLE AS PQR

IN TRIANGLE ABC

AB = 12

BC =?

AC =13

BY PYTHAGORAS THEOREM

${AC^{2}=AB^{2}+BC^{2}$

${13}^{2} = {12}^{2} + {BC}^{2}\\ 169 =144+ BC^{2} \\169-144 = BC^{2}\\BC^{2}=25\\BC = 5$

IN TRIANGLE PQR

PQ =6

QR=?

PR=10

BY PYTHAGORAS THEOREM

${PR^{2}=PQ^{2}+QR^{2}$

$10^{2} = 6^{2} + QR^{2}\\100=36+QR^{2}\\100-36 =QR^2\\ 64 =QR^{2} \\ 8 = QR$

Now,

BC + QR = x

Therefore,

5+8 =x

13=x

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Answer 2

Answer:

• x = 13cm.

Explanation:

Let's name figures as (i), and (ii)

[As shown in the attachment]

Here we've assumed x = p + q

Where, p is base of fig.(i) and q is base of fig.(ii)

In figure (i):-

• Hypotenuse = 13
• perpendicular = 12
• Base = $\rm p$ (assuming)

Using Pythagoras theorem,

$\rm \implies \: {(13)}^{2} = {(12)}^{2} + {(p)}^{2}$

$\rm \implies \: 169 = 144 + {p}^{2}$

$\rm \implies \: 169 - 144 = {p}^{2}$

$\rm \implies \: 25 = {p}^{2}$

$\rm \implies \: \sqrt{25} = {p}$

$\rm \implies \: 5 = {p}$

And also,

In figure (ii):-

• Hypotenuse = 10cm
• Perpendicular = 6cm
• Base = q

Using Pythagoras theorem,

$\rm \implies \: {(10)}^{2} = {(6)}^{2} + {q}^{2}$

$\rm \implies \: 100 = 36 + {q}^{2}$

$\rm \implies \: 100 - 36 = {q}^{2}$

$\rm \implies \: 64 = {q}^{2}$

$\rm \implies \: \sqrt{64} = {q}$

$\rm \implies \: 8 = {q}$

Hence, the value of ‘x’ is :-

$\rm \implies \: x = p + q$

$\rm \implies \: x = 5 + 8$

$\rm \implies \: x = 13$

$\rm \therefore \: x = 13cm$

Required answer: 13cm

__________________________

Formula used:

Pythagoras theorem,

• (h)² = (p)² + (b)²

Where,

• h denotes the hypotenuse.
• p is the perpendicular.
• b is the base.