PE 4 - Triangles | Geometry - Triangles
A rectangle inscribed in a triangle has its base coinciding with the base b of the triangle. If the altitude of the triangle is h, and the altitude x of the rectangle is half the base of the rectangle, then:
- A.
x = h
- B.
x =
- C.
x =
- D.
x =
Answer: Option C
Explanation :
Let ABC be the triangle with h as the altitude and BC as base. Let PQRS be the rectangle inscribed in triangle ABC.
As height of rectangle = PS = x, and
base of the rectangle = SR = 2x = PQ
Now, height of ∆APQ = height of triangle – height of rectangle = h – x
Now, ∆APQ and ∆ABC are similar.
∴ =
∴ =
∴ = 1 -
∴ + = 1
∴ x =
Hence, option (c).
Workspace:
In ∆ABC, CD is the median and E is the mid-point of CD and AE extended meets BC at F. AB, BC and CF are 5, 6 and 7 cm. Find CF.
- A.
3.1
- B.
2.5
- C.
2
- D.
3.45
Answer: Option C
Explanation :
Let us draw DG || AF as shown in the figure.
Since D is the mid-point of AB, AD = DB
Also, since E is the mid-point of CD, CE = ED
In ∆CDG, since DG || EF, hence
CF : FG = CE : ED = 1 : 1
⇒ CF = FG ...(1)
In ∆BAF, since DG || AF, hence
BG : GF = BD : DA = 1 : 1
⇒ BG = GF ...(2)
From (1) and (2), we get
BG = GF = FC
∴ BG = GF = FC = ⅓ BC = 2 cm.
Hence, option (c).
Workspace:
Area of triangle S1 is 36 cm2. Another triangle S2 is made by joining mid-points of S1. Another triangle S3 is made by joining mid-points of S2. This process is repeated indefinitely. What will be the sum of area of all such triangles formed.
- A.
48
- B.
72
- C.
60
- D.
Cannot be determined
Answer: Option A
Explanation :
We know area of smaller triangle formed by joining mid-points of a triangle is one-fourth of bigger triangle.
∴ Area of S2 = ¼ (Area of S1) = 36/4
Area of S3 = ¼ (Area of S2) = 36/42
Area of S4 = ¼ (Area of S3) = 36/43
... and so on
∴ Sum of all the areas = 36 + + + + ...
This is sum of an infinite GP whose first term is 36 and common ratio = 1/4
∴ Sum of all the areas = = = 48
Hence, option (a).
Workspace:
In the figure given below EF || BC, AE = BE and DG = GE. In area of ∆AEF = 10 cm2, find the area of ∆DGH.
- A.
10
- B.
5
- C.
15
- D.
12.5
- E.
Cannot be determined
Answer: Option B
Explanation :
Let us draw perpendiculars from D to EF and A to BC as shown in the figure.
Since EF || BC and AE = EB, it means
⇒ E and F are the mid-points of AB and AC respectively.
⇒ Also, AS = SR ...(1)
Also EF || GH and DG = GE, it means
⇒ G and H are the mid-points of DE and DF respectively.
⇒ GH = ½ EF ...(2)
⇒ Also, DQ = PQ ...(3)
From (1) and (3), we get
AS = SR = PQ = DQ
⇒ AS = DQ
In ∆AEF and ∆DGH,
perpendicular is same i.e., AS = DQ,
while base GH of ∆DGH is half the base EF of ∆AEF
∴ Area of ∆DGH will be half the area of ∆AEF = ½ × 10 = 5 cm2.
Hence, option (b).
Workspace:
In an isosceles triangle ∆ABC, AB = AC. It is folded along the altitude AD such that AB and AC overlap. The triangle obtained is also an isosceles triangle whose area is 8 cm2. Find AB.
- A.
2√2
- B.
8
- C.
4
- D.
4√2
Answer: Option D
Explanation :
As we can see from the figure above, the folded triangle obtained will be a right isosceles triangle.
⇒ Area of folded triangle = 8 = ½ × a × a
⇒ a = 4
∴ Height of the original triangle = 4, and
Base of the original triangle = 2a = 8
Now, in ∆ABD,
AB2 = AD2 + BD2
⇒ AB2 = 16 + 16 = 32
⇒ AB = 4√2
Hence, option (d).
Workspace:
The altitudes of triangle are 12,15 and 20 units. Find the area (in sq. units) of the triangle.
[Type in your answer as the nearest possible integer]
Answer: 150
Explanation :
Let the base side for altitude 12, 15 and 20 be a, b and c respectively.
∴ Area of triangle = ½ × 12 × a = ½ × 15 × b = ½ × 20 × c
⇒ 12a = 15b = 20c
⇒ a : b : c = 5 : 4 : 3
∴ a = 5k, b = 4k and c = 3k
Now, b2 + c2 = a2,
∴ The given triangle is a right triangle.
⇒ The perpendicular on two smaller sides are equal to sides.
The two smaller sides will be 15 and 20
∴ Area of triangle = ½ × 15 × 20 = 150.
Hence, 150.
Workspace:
If the altitudes of a triangle are 3, 4 and 6. Find the inradius of this triangle.
- A.
3/2
- B.
2/3
- C.
4/3
- D.
3/4
- E.
Cannot be determined
Answer: Option C
Explanation :
Let the base sides for altitudes 3, 4 and 6 be a, b and c respectively.
∴ Area of triangle (∆) = ½ × 3 × a = ½ × 4 × b = ½ × 6 × c
⇒ ∆ = ½ × 3 × a
⇒ = ...(1)
⇒ ∆ = ½ × 4 × b
⇒ = ...(2)
⇒ ∆ = ½ × 6 × c
⇒ = ...(3)
Adding (1), (2) and (3), we get
= + + =
⇒ =
⇒ =
⇒ =
⇒ r =
Hence, option (c).
Workspace:
In the triangle given, BD : DC = 2 : 3, AF : FC = 3 : 4. Find BE : EF.
- A.
9 : 14
- B.
14 : 9
- C.
8 : 15
- D.
8 : 15
- E.
Cannot be determined
Answer: Option B
Explanation :
In ∆ABD and ∆ACD, since height is same, the ratio of areas will be same as the ratio of bases.
⇒ = =
Now, in ∆BED and ∆CED, since height is same, the ratio of areas will be same as the ratio of bases.
⇒ = =
Now, = =
∴ = = =
⇒ =
Let Area(∆ABE) = 2A and Area(∆ACE) = 3A ...(1)
Now, since AF : FC = 3 : 4
⇒ =
From (1),
Area(∆AEF) = 3A × 3/7 = 9A/7 and Area(∆ACE) = 3A × 4/7 = 12A/7 ...(2)
From (1) and (2),
= = =
Hence, option (b).
Workspace:
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