Question
Find $AB.$
✓
Answer
Consider the figure
From right $\triangle ACF$
$\tan 45^{\circ}=\frac{20}{ AC } $
$1=\frac{20}{ AC } $
$AC =20 \ cm$
From $\triangle DEB$
$\tan 60^{\circ}=\frac{30}{ BD }$
$\sqrt{3}=\frac{30}{ BD }$
$BD =\frac{30}{\sqrt{3}}=17.32 \ cm$
Given $FC = 20, ED = 30,$
So $EP = 10 \ cm$
Therefore
$\tan 60^{\circ}=\frac{ FP }{ EP }$
$\sqrt{3}=\frac{ FP }{10}$
$FP =10 \sqrt{3}=17.32 \ cm $
Thus $AB = AC + CD + BD = 54.64 \ cm$
From right $\triangle ACF$
$\tan 45^{\circ}=\frac{20}{ AC } $
$1=\frac{20}{ AC } $
$AC =20 \ cm$
From $\triangle DEB$
$\tan 60^{\circ}=\frac{30}{ BD }$
$\sqrt{3}=\frac{30}{ BD }$
$BD =\frac{30}{\sqrt{3}}=17.32 \ cm$
Given $FC = 20, ED = 30,$
So $EP = 10 \ cm$
Therefore
$\tan 60^{\circ}=\frac{ FP }{ EP }$
$\sqrt{3}=\frac{ FP }{10}$
$FP =10 \sqrt{3}=17.32 \ cm $
Thus $AB = AC + CD + BD = 54.64 \ cm$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
More "[5 marks sum]" from Solution of Right Triangles [Simple 2-D Problems Involving One Right-angled Triangle]→Solution of Right Triangles [Simple 2-D Problems Involving One Right-angled Triangle] — all question types→MATHEMATICS STD 9 question papers→ICSE Board STD 9 question papers→ICSE Board question paper generator→
Similar questions
Evaluate the following:$\left(\frac{64}{216}\right)^{\frac{2}{3}} \times\left(\frac{16}{36}\right)^{-\frac{3}{2}}$
→A man lends $Rs. 12,500$ at $12\ \%\ $ for the first year, at $15\ \%\ $ for the second year and at $18\ \%\ $ for the third year. If the rates of interest are compounded yearly ; find the difference between the $C.I.$ fo the first year and the compound interest for the third year.
→Case Study : In 2015, the population of a town was 20000. During 2016 and 2017, it increases by 4% and 5% every year respectively. In 2018, due to an epidemic and migration to cities, the population decreases at 5%.
Based on the above information answer the following questions:
Q.1. The population of the town in 2016 was:
(a) 20800 (b) 20500 (c) 20460 (d) 20300
Q.2. The difference in the population of the town at the end of the years 2017 and 2015 was:
(a) 1500 (b) 1700 (e) 1800 (d) 1840
Q.3. The population of the town at the end of the year 2018 was:
(a) 20100 (b) 20500 (c) 20748 (d) 20850
Q.4. Which of the following expressions gives the population at the end of the year 2018?
(a) $20000 \quad \frac{104}{100} \quad \frac{105}{100} \quad \frac{105}{100}$
(b) $20000 \quad \frac{104}{100} \quad \frac{105}{100} \quad \frac{95}{100}$
(c) $20000 \quad \frac{104}{100} \quad \frac{95}{100} \quad \frac{95}{100}$
(d) $20000 \quad \frac{96}{100} \quad \frac{95}{100} \quad \frac{95}{100}$
Q.5. If the population of the town at the end of 2019 was 21163, then during 2019, the population increases at the rate of:
(a) 2% (b) 3% (c) 3.5% (d) 4%
→Based on the above information answer the following questions:
Q.1. The population of the town in 2016 was:
(a) 20800 (b) 20500 (c) 20460 (d) 20300
Q.2. The difference in the population of the town at the end of the years 2017 and 2015 was:
(a) 1500 (b) 1700 (e) 1800 (d) 1840
Q.3. The population of the town at the end of the year 2018 was:
(a) 20100 (b) 20500 (c) 20748 (d) 20850
Q.4. Which of the following expressions gives the population at the end of the year 2018?
(a) $20000 \quad \frac{104}{100} \quad \frac{105}{100} \quad \frac{105}{100}$
(b) $20000 \quad \frac{104}{100} \quad \frac{105}{100} \quad \frac{95}{100}$
(c) $20000 \quad \frac{104}{100} \quad \frac{95}{100} \quad \frac{95}{100}$
(d) $20000 \quad \frac{96}{100} \quad \frac{95}{100} \quad \frac{95}{100}$
Q.5. If the population of the town at the end of 2019 was 21163, then during 2019, the population increases at the rate of:
(a) 2% (b) 3% (c) 3.5% (d) 4%
Find the base of an isosceles triangle whose area is $192\ cm^2$ and the length of one of the equal sides is $20\ cm.$
→In the given figure, ABCD is a trapezium in which $AD =13 cm, BC =5 cm, CD =17 cm$ and $\angle A =\angle B =90$.
Calculate :
(i) AB
(ii) Area of trap. ABCD .
→Calculate :
(i) AB
(ii) Area of trap. ABCD .
Solve for $x:2^{x+3}+2^{x+1}=320$
→Plot each of the following points on a graph paper.
(i) A(6, 3) (ii) B(- 4, 1) (iii) C(- 2,- 5) (iv) D(2, - 5)
(v) P(4,0) (vi) Q(0, 3) (vii) R(- 3, - 3) (viii) S(0, - 3)
(i) A(6, 3) (ii) B(- 4, 1) (iii) C(- 2,- 5) (iv) D(2, - 5)
(v) P(4,0) (vi) Q(0, 3) (vii) R(- 3, - 3) (viii) S(0, - 3)
In $\triangle P Q R, P S$ is a median. $T$ is the mid$-$point of $S R$ and $M$ is the mid$-$point of $P T$. Prove that: $\triangle P M R=\frac{1}{8} \Delta PQR$
→A rectangular tank $30 \ cm \times 20 \ cm \times 12 \ cm$ contains water to a depth of $6 \ cm$ . A metal cube of side $10 \ cm$ is placed in the tank with its one face resting on the bottom of the tank. Find the volume of water, in liters, that must be poured in the tank so that the metal cube is just submerged in the water.
→A trader fixes the selling price of his goods at $50\%$ above the cost price. He sells half of his stock at this price, a quarter of his stock at a discount of $20\%$ on the original selling price, and the rest at a discount of $36\%$ on the original selling price. Find the gain per cent altogether.
→