1 Crore+ students have signed up on EduRev. Have you? 
Two poles of height 9 m and 14 m stand on a plane ground. If the distance between their feet is 12 m, then the distance between their tops is
Let AB and CD be the heights of given poles, BC be the distance between their feet and AD be the distance between their tops
Since, BCDE is a rectangle
∴ EB = DC = 9 m and ED = BC = 12 m
AE = AB – EB = 14 m – 9 m = 5m
Now, ΔAED is a right angled triangle
∴ AD^{2} = AE^{2} + ED^{2} [By Pythagoras theorem]
⇒ AD^{2} = (5)^{2} + (12)^{2} = 169 ⇒ AD = 13 m
Hence, the distance between their tops is 13 m.
If a tree casts a 18 feet shadow and at the same time, a child of height 3 feet casts a 2 feet shadow, then the height of the tree is
Let AB be the height of tree which casts a shadow AC = 18 feet and ED be the height of child which casts a shadow CD = 2 feet
Now, in ΔABC and ΔDEC, we have
∠A = ∠D [Each 90°]
∠C = ∠C [Common angle]
∴ ΔABC ~ ΔDEC [By AA similarity]
⇒ AB/DE = AC/DC
⇒ AB/3 = 18/2
⇒ AB = 27 feet
So, the height of tree is 27 feet.
An aeroplane leaves an airport and flies due north at a speed of 1000 km per hour. At the same time, another aeroplane leaves the same airport and flies due west at a speed of 1200 km per hour. How far apart will be the two planes after 1(1/2) hours?
Let the point A represent the airport.
Let the planeI fly towards North.
∴ Distance of the planeI from the airport after 1(1/2) hours
= Speed × Time = 1000 × (1)1/2 km= 1500 km
Let the planeII fly towards West.
∴ Distance of the planeII from the airport after (1)1/2 hours
= 1200 × (1)1/2km = 1800 km
Now, in right ∆ABC, using Pythagoras theorem, we have
BC^{2} = AB^{2 }+ AC^{2}
⇒ BC^{2} = (1800)^{2} + (1500)^{2} = 5490000
⇒ BC =
Thus, after (1)1/2 hours, the two planes will be 300 √61 km apart from each other.
A 12 cm rod is held between a flashlight and a wall as shown. Find the length of the shadow on the wall if the rod is 45 cm from the wall and 15 cm from the light.
In ΔADM and ΔABF
∠DAM = ∠BAF (Common)
∠DMA = ∠BFM (Corresponding angles)
∴ ΔADM ~ ΔABF (By AA similarity)
∴ BC = BF + FC = 2BF = 48 cm (∵ CF = BF)
In the given figure, ABC is a right triangle rightangled at B. AD and CE are the two medians drawn from A and C respectively. If AC = 5 cm and AD = then the length of CE is
Since ΔABD is a right angled triangle
∴ AD^{2} = AB^{2} + BD^{2} [By using Pythagoras theorem]
⇒ AD^{2} = AB^{2} + (BC/2)^{2} [∵ AD is median on BC]
⇒ AD^{2} = AB^{2} + (1/4) BC^{2} ...(i)
Similarly in, ΔBCE
CE^{2} = BC^{2 }+ BE^{2} ⇒ CE^{2 }= BC^{2} + (AB/2)^{2 }[∵ EC is median on AB]
⇒ CE^{2} = BC^{2} + (1/4).AB^{2} ...(ii)
Adding (i) and (ii), we get
Mason Construction wants to connect two parks on opposite sides of town with a road. Surveyors have laid out a map as shown. The road can be built through the town or around town through point R. The roads intersect at a right angle at point R. The line joining Park A to Park B is parallel to the line joining C and D.
(i) What is the distance between the parks through town?
(ii) What is the distance from Park A to Park B through point R?
(i) Since, CD  AB [Given]
∴ In ΔRCD and ΔRBA, we have
∠BAR = ∠RDC [Alternate interior angles]
∠ABR = ∠RCD
∴ ΔRCD ~ ΔRBA [By AA similarity]
... (i)
i.e., Distance between the parks through town is 8.75 m.
(ii) In right ΔCRD, we have
(CD)^{2 }= (CR)^{2} + (RD)^{2}
⇒ RD^{2} = (1.4)^{2} – (1.2)^{2} = 0.52 ⇒ RD = 0.72 m
Since ΔRCD ~ ΔRBA
i.e., Distance from Park A to Park B through point R = AR + RB
= 4.5 m + 7.5 m = 12 m
BPT (Basic proportional theorem) states that if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points then the other two sides are divided in the present ratio.
(P) Given :
∴ ∠A is containing the sides AB and AC and ∠P is containing the sides PQ and PR.
∴ ΔABC ~ ΔPQR (By SAS criteria)
(Q) Given : ∠A = ∠P, ∠B = ∠Q
∴ ΔABC ~ ΔPQR (By AA criteria)
∵ Sides of the ΔABC and ΔPQR are in proportion
∴ ΔABC ~ ΔPQR (By SSS criteria)
(S) Given, DE  BC
In the given figure,the line segment XY is parallel to side AC of ΔABC and it divides the triangle into two parts of equal area. Then, find
(i) AX : AB (ii) AC/XY
(i) Since, XY  AC, we have
∠A = ∠BXY and ∠C = ∠BYX [Corresponding angles]
∴ ΔABC ~ ΔXBY [By AA similarity]
...(i)
But, ar(ΔABC) = 2 × ar(ΔXBY) [Given]
...(ii)
From (i) and (ii), we get
Hence, AX : AB = (2 − √2) : 2
(ii) Since, ΔABC ~ ΔXBY
P and Q are the midpoints of the sides CA and CB respectively of a DABC, right angled at C, then find:
(i) 4AC^{2} + BC^{2}
(ii) 4BC^{2} + AC^{2}
(iii) 4(AQ^{2} + BP^{2})
Construction: Join PQ, BP and AQ.
(i) 4AC^{2} + BC^{2} = 4AC^{2} + (2QC)^{2}
[∵ Q is midpoint of BC]
= 4AC^{2} + 4QC^{2}
= 4(AC^{2} + QC^{2})
= 4AQ^{2} [∵ AQC is right angled triangle]
(ii) 4BC^{2} + AC^{2} = 4BC^{2} + (2CP)^{2} [∵ P is midpoint of AC]
= 4BC^{2} + 4CP^{2} = 4(BC^{2} + CP^{2}) = 4BP2 [∵ PBC is right angled triangle]
(iii) 4[AC^{2} + QC^{2}] = 4(AQ)^{2} [from (i) part]
⇒ 4AC^{2} + BC^{2} = 4AQ^{2} ...(i)
4[BC^{2} + CP^{2}] = 4(BP)^{2} [from (ii) part]
⇒ 4BC^{2} + AC^{2} = 4BP^{2} ...(ii)
Now, adding (i) and (ii), we get
4(AQ^{2} + BP^{2}) = 5(AC^{2} + BC^{2}) = 5AB^{2} [∵ ABC is right angled triangle]
11 videos36 docs201 tests

Use Code STAYHOME200 and get INR 200 additional OFF

Use Coupon Code 
11 videos36 docs201 tests









