Question : $\triangle ABC$ is an isosceles triangle with AB = AC = 15 cm and an altitude from A to BC of 12 cm. The length of side BC is:
Option 1: 9 cm
Option 2: 12 cm
Option 3: 18 cm
Option 4: 20 cm
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Correct Answer: 18 cm
Solution : Given: $\triangle ABC$ is an isosceles triangle with $AB = AC = 15$ cm. $AD \perp BC$ So, $AD = 12$ cm. Also, $BD = DC$. Using Pythagoras theorem in $\triangle ABD$, $(BD)^2=(AB)^2–(AD)^2$ $⇒BD=\sqrt{(AB)^2–(AD)^2}$ $⇒BD=\sqrt{(15)^2–(12)^2}$ $⇒BD=\sqrt{(15–12)(15+12)}$ $⇒BD=\sqrt{3\times27}$ $⇒BD=9$ cm $BC = 2 \times BD = 2 \times 9 =18$ cm. So, the length of the side BC is 18 cm. Hence, the correct answer is 18 cm.
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Question : ABC is an isosceles triangle inscribed in a circle. If AB = AC = $12\sqrt{5}$ cm and BC = 24 cm, then the radius of circle is:
Option 1: 10 cm
Option 2: 15 cm
Option 3: 12 cm
Option 4: 14 cm
Question : If $\triangle ABC \sim \triangle PQR$, AB =4 cm, PQ=6 cm, QR=9 cm and RP =12 cm, then find the perimeter of $\triangle$ ABC.
Option 1: 18 cm
Option 2: 16 cm
Option 3: 20 cm
Option 4: 22 cm
Question : Suppose $\triangle ABC$ be a right-angled triangle where $\angle A=90°$ and $AD\perp BC$. If the area of $\triangle ABC =40$ cm$^{2}$ and $\triangle ACD =10$ cm$^{2}$ and $\overline{AC}=9$ cm, then the length of $BC$ is:
Option 1: 12 cm
Option 2: 18 cm
Option 3: 4 cm
Option 4: 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 : Triangle ABC and DEF are similar. If AB = 92 cm, BC = 48 cm, AC =120 cm, and the length of the smallest side of triangle DEF is 200 cm, then find the length of the longest side of triangle DEF.
Option 1: 400 cm
Option 2: 225 cm
Option 3: 350 cm
Option 4: 500 cm
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