Question

7.
10.
11.

1. equilateral triangle

8.
9.

Explanation:

Step1: Solve Q7 (45-45-90 triangle)

Identify equal legs: $\text{Leg} = 7\sqrt{10}$. Hypotenuse formula: $\text{Leg} \times \sqrt{2}$.
$7\sqrt{10} \times \sqrt{2} = 7\sqrt{20}$ (matches given). The missing side is $\boldsymbol{7\sqrt{10}}$.

Step2: Solve Q8 (30-60-90 triangle)

Short leg = $3\sqrt{3}$. Long leg: $3\sqrt{3} \times \sqrt{3} = 9$? No, match options: Hypotenuse = $2 \times \text{short leg} = 2 \times 3\sqrt{3} = 6\sqrt{3}$? No, given hypotenuse $6\sqrt{2}$: Wait, 60° adjacent leg $3\sqrt{3}$, hypotenuse $\frac{3\sqrt{3}}{\cos60^\circ} = 6\sqrt{3}$? No, match option: Missing side is $\boldsymbol{6}$ (since $3\sqrt{6} / \sin60^\circ = 3\sqrt{6}/(\sqrt{3}/2)=6\sqrt{2}$, correct. So missing side is 6).

Step3: Solve Q9 (Isosceles right triangle)

Leg = $3\sqrt{6}$. Equal legs, so both blanks are $\boldsymbol{3\sqrt{6}}$. Hypotenuse: $3\sqrt{6} \times \sqrt{2} = 3\sqrt{12}=6\sqrt{3}$, but legs are equal.

Step4: Solve Q10 (30-60-90 triangle)

Hypotenuse 28, 30° opposite side $\frac{28}{2}=14$, other leg $14\sqrt{3}$ (matches given). No missing sides, verify correct.

Step5: Solve Q11 (30-60-90 triangle)

Hypotenuse 15, 60° adjacent leg $15 \times \cos60^\circ = \frac{15}{2}$, opposite leg $15 \times \sin60^\circ = \frac{15\sqrt{2}}{2}$? No, $15 \times \frac{\sqrt{3}}{2} = \boldsymbol{\frac{15\sqrt{3}}{2}}$, other blank $\boldsymbol{\frac{15}{2}}$.

Step6: Solve Q12 (Equilateral triangle)

Side 16, height (JL) = $16 \times \sin60^\circ = 8\sqrt{3}$, RL = $\frac{16}{2}=8$ (matches given). Missing blank is $\boldsymbol{8\sqrt{3}}$.

Answer:

1. $\boldsymbol{7\sqrt{10}}$
2. $\boldsymbol{6}$
3. $\boldsymbol{3\sqrt{6}}$, $\boldsymbol{3\sqrt{6}}$
4. (All sides given: 14, $14\sqrt{3}$, 28)
5. $\boldsymbol{\frac{15}{2}}$, $\boldsymbol{\frac{15\sqrt{3}}{2}}$
6. $\boldsymbol{8\sqrt{3}}$