[3 marks sum]

Question 13 Marks

In triangle PQR, angle Q = 90°, find: PR, if PQ = 8 cm and QR = 6 cm

Answer

Given:
$
\begin{aligned}
& PQ =8 cm \\
& QR =6 cm \\
& PR =? \\
& \angle PQR =90^{\circ}
\end{aligned}
$

According to Pythagoras Theorem,
$
\begin{aligned}
& (P R)^2=(P Q)^2+(Q R)^2 \\
& P R^2=8^2+6^2 \\
& P R^2=64+36 \\
& P R^2=100 \\
& \therefore P R=\sqrt{100}=10 cm
\end{aligned}
$

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Question 23 Marks

In the given figure, angle $B A C=90^{\circ}, A C=400 m$, and $A B=300 m$. Find the length of $B C$.

Answer

$\begin{aligned} & A C=400 m \\ & A B=300 m \\ & B C=?\end{aligned}$

According to Pythagoras Theorem,
$
\begin{aligned}
& B C^2=A B^2+A C^2 \\
& B C^2=(300)^2+(400)^2 \\
& B C^2=90000+160000 \\
& B C^2=250000 \\
& B C=\sqrt{250000}=500 m
\end{aligned}
$

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Question 33 Marks

The sides of a certain triangle is given below. Find, which of them is right-triangle
$6 m, 9 m \text {, and } 13 m$

Answer

$6 m, 9 m$, and $13 m$
The given triangle will be a right-angled triangle if the square of its largest side is equal to the sum of the squares on the other two sides.
i.e., If $(13)^2 = (9)^2 + (6)^2$
$169 = 81 + 36 = 169 \neq 117$
So, the given triangle is not right-angled.

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Question 43 Marks

The sides of a certain triangle is given below. Find, which of them is right-triangle
$16 cm, 20 cm$ and $12 cm$

Answer

$16 cm, 20 cm$ and $12 cm$
The given triangle will be a right-angled triangle if square of its largest side is equal to the sum of the squares on the other two sides.
i.e., If $(20)^2 = (16)^2 = (12)^2$
$(20)^2 = (16)^2 + (12)^2$
$400 = 256 + 144$
$400 = 400$
So, the given triangle is right-angled.

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Question 53 Marks

Triangle XYZ is right-angled at vertex Z. Calculate the length of YZ, if XY = 13 cm and XZ = 12 cm.

Answer

Given: $\triangle X Y Z$ right angled at $Z$ and $X Y=13 cm , X Z=12 cm$.
To find: Length of YZ.
According to Pythagoras Theorem,
$
\begin{aligned}
& X Y^2=X Z^2+Y Z^2 \\
& 13^2=12^2+Y Z^2
\end{aligned}
$

$\begin{aligned} & 169=144+ YZ ^2 \\ & 169-144= YZ ^2 \\ & 25=Y Z^2 \\ & \therefore Y Z=\sqrt{25} cm =\sqrt{5 \times 5}=5 cm \end{aligned}$

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Question 63 Marks

Use the information given in the figure to find the length $A D$.

Answer

Given:
$ \begin{aligned} & A B=20 \mathrm{~cm} \\ & \therefore A O=A B=\frac{20}{2}=10 \mathrm{~cm} \\ & B C=O D=24 \mathrm{~cm} \end{aligned} $
To find: Length of $A D$
In right angled triangle
$ \begin{aligned} & A O D(A D)^2=(A O)^2+(O D)^2 \\ & (A D)^2=(10)^2+(24)^2 \\ & (A D)^2=100+576 \\ & (A D)^2=676 \\ & \therefore A D=\sqrt{26 \times 26} \\ & A D=26 \mathrm{~cm} \end{aligned} $

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Question 73 Marks

Triangle ABC is right-angled at vertex A. Calculate the length of BC, if AB = 18 cm and AC = 24 cm.

Answer

Given: $\triangle A B C$ right angled at $A$ and $A B=18 cm , A C=24 cm$.

To find: Length of $B C$. According to Pythagoras Theorem,
$
\begin{aligned}
& B C^2=A B^2+A C^2 \\
& =18^2+24^2=324+576=900 \\
& \therefore B C=\sqrt{900}=\sqrt{30 \times 30}=30 cm
\end{aligned}
$

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MATHS [3 marks sum]

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