Question

fill in the blanks with the missing lengths
1.
2.
3.
4.
5.
6.
options: 4√10, 7, 8, 8√2, 8√5, 11, 12√2, 13, 14

Explanation:

Step1: Solve Q1 (45-45-90 triangle)

In a 45-45-90 triangle, legs are equal, hypotenuse = leg$\times\sqrt{2}$.
Leg = 8, so hypotenuse = $8\sqrt{2}$, missing leg = 8.

Step2: Solve Q2 (45-45-90 triangle)

Hypotenuse = $13\sqrt{2}$, so legs = $\frac{13\sqrt{2}}{\sqrt{2}}=13$.

Step3: Solve Q3 (45-45-90 triangle)

Leg = $4\sqrt{10}$, so other leg = $4\sqrt{10}$, hypotenuse = $4\sqrt{10}\times\sqrt{2}=4\sqrt{20}=8\sqrt{5}$.

Step4: Solve Q4 (Right triangle, Pythagoras)

Let missing side = $x$. Use $a^2+b^2=c^2$:
$x^2 + (\sqrt{21})^2 = 10^2$
$x^2 +21=100$
$x^2=79$? No, wait: $10^2 - (\sqrt{21})^2=100-21=79$? No, wait the options have 11: $11^2 + (\sqrt{21})^2=121+21=142≠100$. Wait, 10 is leg, hypotenuse is missing: $10^2+(\sqrt{21})^2=100+21=121=11^2$. Yes, hypotenuse=11.

Step5: Solve Q5 (Square diagonal, 45-45-90)

Diagonal of square = 24, side = $\frac{24}{\sqrt{2}}=12\sqrt{2}$.

Step6: Solve Q6 (30-60-90 triangle)

Long leg = $7\sqrt{3}$, so short leg = 7, hypotenuse = $2\times7=14$.

Answer:

1. Missing leg: 8, Hypotenuse: $8\sqrt{2}$
2. Missing legs: 13, 13
3. Missing leg: $4\sqrt{10}$, Hypotenuse: $8\sqrt{5}$
4. Missing hypotenuse: 11
5. Missing side: $12\sqrt{2}$
6. Missing short leg: 7, Hypotenuse: 14