Question : $ABCD$ is a parallelogram in which $AB = 7 \operatorname{cm}, BC = 9 \operatorname{cm}$, and diagonal $AC = 8 \operatorname{cm}$. What is the length (in cm) of the other diagonal?
Option 1: $14$
Option 2: $14\sqrt{2}$
Option 3: $7$
Option 4: $7\sqrt{2}$
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Correct Answer: $14$
Solution : Given: AC = 8 cm, AB = 7 cm, and BC = 9 cm In parallelogram $ABCD$, The formula of diagonals is $AC^2 + BD^2 = 2(AB^2 + BC^2)$ $\Rightarrow 8^2 + BD^2 = 2(7^2 + 9^2)\\$ $\Rightarrow 64 + BD^2 = 2(49 + 81) = 260\\$ $\Rightarrow BD^2 = 260 - 64 = 196\\$ $\Rightarrow BD = 14 \text{ cm}\\$ Hence, the correct answer is $14$ cm.
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Question : The diagonal of the square is $8 \sqrt{2}$ cm. Find the diagonal of another square whose area is triple that of the first square.
Option 1: $8 \sqrt{5}$ cm
Option 2: $8 \sqrt{3}$ cm
Option 3: $8 \sqrt{2}$ cm
Option 4: $8 \sqrt{6}$ cm
Question : In a triangle ${ABC}, {AB}={AC}$ and the perimeter of $\triangle {ABC}$ is $8(2+\sqrt{2}) $ cm. If the length of ${BC}$ is $\sqrt{2}$ times the length of ${AB}$, then find the area of $\triangle {ABC}$.
Option 1: $32 \ cm^2$
Option 2: $28 \ cm^2$
Option 3: $16 \ cm^2$
Option 4: $36 \ cm^2$
Question : $AB$ and $AC$ are the two tangents to a circle whose radius is 6 cm. If $\angle BAC=60^{\circ}$, then what is the value (in cm) of $\sqrt{\left ( AB^{2}+AC^{2} \right )}?$
Option 1: $6\sqrt{6}$
Option 2: $4\sqrt{6}$
Option 3: $9\sqrt{3}$
Option 4: $8\sqrt{3}$
Question : The area of the parallelogram whose length is 30 cm, width is 20 cm and one diagonal is 40 cm is:
Option 1: $200\sqrt{15}\text{ cm}^{2}$
Option 2: $100\sqrt{15}\text{ cm}^{2}$
Option 3: $300\sqrt{15}\text{ cm}^{2}$
Option 4: $150\sqrt{15}\text{ cm}^{2}$
Question : In $\triangle$LMN, LM = $5\sqrt{2}$ cm, LN = 13 cm and $\angle$LMN=135$^{\circ}$. What is the length (in cm) of MN?
Option 1: $7$
Option 2: $8$
Option 3: $8\sqrt{2}$
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