Solutions: Triangles - Test 3

Q1. (b) 7 cm, 24 cm, 25 cm

$7^2 + 24^2 = 49 + 576 = 625 = 25^2$.

Q2. (c) $90^\circ$

$AB^2 = 108, BC^2 = 36, AC^2 = 144$. $AB^2 + BC^2 = AC^2$. By converse of Pythagoras theorem, $\angle B = 90^\circ$.

Q3. (a) 10 cm

Diagonal $d = a\sqrt{2} \Rightarrow 10\sqrt{2} = a\sqrt{2} \Rightarrow a = 10$.

Q4. (b) 6 m

Base $b = \sqrt{10^2 - 8^2} = \sqrt{100 - 64} = \sqrt{36} = 6$.

Q5. (a) Both A and R are true and R is the correct explanation of A.

The statement A is the definition of Pythagoras Theorem.

Q6. Difference in height = $11 - 6 = 5$ m. Horizontal distance = 12 m.
Distance between tops = $\sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13$ m. [2]

Q7. In $\Delta ABC$, $\angle C = 90^\circ$ and $AC = BC$ (Isosceles).
$AB^2 = AC^2 + BC^2 = AC^2 + AC^2 = 2AC^2$. [2]

Q8. Altitude $h$ bisects the base $2a$ into $a$.
$h^2 + a^2 = (2a)^2 \Rightarrow h^2 = 4a^2 - a^2 = 3a^2 \Rightarrow h = a\sqrt{3}$. [2]

Q9. Let diagonals of rhombus $ABCD$ intersect at $O$. $AC \perp BD$.
In $\Delta AOB$, $AB^2 = OA^2 + OB^2 = (\frac{AC}{2})^2 + (\frac{BD}{2})^2$.
$AB^2 = \frac{AC^2}{4} + \frac{BD^2}{4} \Rightarrow 4AB^2 = AC^2 + BD^2$.
Since $AB=BC=CD=DA$, $AB^2+BC^2+CD^2+DA^2 = AC^2 + BD^2$. [3]

Q10. Draw $AM \perp BC$. $BM = MC = BC/2 = AB/2$.
$DM = BM - BD = \frac{AB}{2} - \frac{AB}{3} = \frac{AB}{6}$.
In $\Delta ADM$, $AD^2 = AM^2 + DM^2$. In $\Delta ABM$, $AM^2 = AB^2 - BM^2$.
$AD^2 = AB^2 - (\frac{AB}{2})^2 + (\frac{AB}{6})^2 = AB^2 - \frac{AB^2}{4} + \frac{AB^2}{36}$.
$36AD^2 = 36AB^2 - 9AB^2 + AB^2 = 28AB^2 \Rightarrow 9AD^2 = 7AB^2$. [3]

(i) Distance = Speed $\times$ Time = $1000 \times 1.5 = 1500$ km. [1]

(ii) Distance = $1200 \times 1.5 = 1800$ km. [1]

(iii) Distance apart = $\sqrt{1500^2 + 1800^2} = 300\sqrt{5^2 + 6^2} = 300\sqrt{25+36} = 300\sqrt{61}$ km. [2]

(i) Distance = $\sqrt{24^2 - 18^2} = \sqrt{576 - 324} = \sqrt{252} = \sqrt{36 \times 7} = 6\sqrt{7}$ m. [2]

(ii) Distance = $\sqrt{30^2 - 18^2} = \sqrt{900 - 324} = \sqrt{576} = 24$ m. [1]

(iii) Pythagoras Theorem. [1]