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Maths 1 Marks Question

P-2 Pythagoras Theorem · Maharashtra Board STD 10 (MSBSHSE - English Medium). 8 questions.

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Question 11 Mark
Find the diagonal of a rectangle whose length is $35 cm$ and breadth is $12\ cm.$
 
Answer
The diagonal $=\sqrt{ }\left[\right.$ length ${ }^2+$ breadth $\left.^2\right]$
$=\sqrt{ }\left(35^2+12^2\right)$
$=\sqrt{ }(1225+144)$
$=\sqrt{ } 1369$
$=37$
Thus the diagonal is $37\ cm .$
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Question 51 Mark
Find the diagonal of a rectangle whose length is 35 cm and breadth is 12 cm.
Answer
The diagonal = √[length2 + breadth2]
= √(352 + 122)
= √(1225 + 144)= √1369
= 37
Thus the diagonal is 37 cm.
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Question 61 Mark
In ∆ LMN ,l = 5, m = 13, n = 12. State whether ∆LMN is a right angled triangle or not.
Answer

$\begin{array}{l}l=5, m=13, \quad n=12 \\
l^2=25, \quad m^2=169, \quad n^2=144 \\
\therefore m^2=l^2+n^2
\end{array}$
$\therefore$ by converse of Pythagoras theorem $\Delta LMN$ is a right angled triangle.
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Question 71 Mark
See fig, In ∆ ABC, seg AD ⊥ seg BC, ∠C = 45°, BD = 5 and AC = 8√2 then find AD and BC.
Image
Answer

$\begin{array}{l}\text { In } \triangle A D C \\ \angle A D C=90^{\circ}, \angle C=45^{\circ}, \therefore \angle D A C=45^{\circ} \\ A D=D C=\frac{1}{\sqrt{2}} \times 8 \sqrt{2} \ldots \text { by } 45^{\circ}-45^{\circ}-90^{\circ} \text { theorem } \\ D C=8 \quad \therefore A D=8 \\ B C=B D+D C \\ \quad=5+8 \\ B C=13\end{array}$
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Question 81 Mark
In fig 2.13, ∠ PQR = 90°,seg QN ⊥ seg PR, PN = 9, NR = 16. Find QN.
Image
Answer
In $\triangle PQR$, seg $QN \perp$ seg $PR$
$NQ ^2= PN \times NR \ldots$ theorem of geometric mean
$\begin{aligned}
\therefore NQ & =\sqrt{ PN \times NR } \\
& =\sqrt{9 \times 16} \\
& =3 \times 4 \\
& =12
\end{aligned}$
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