CAT — Quadrilaterals
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ABCD is a trapezium in which AB is parallel to CD. The sides AD and BC when extended, intersect at point E. If AB = 2 cm, CD = 1 cm, and perimeter of ABCD is 6 cm, then the perimeter, in cm, of △AEB is
8
10
9
7
8
Given:
ABCD is a trapezium with
AB ∥ CD
AB = 2 cm
CD = 1 cm
Perimeter of ABCD = 6 cm.
The extensions of AD and BC meet at E.
Find the perimeter of △AEB.
Step 1: Use similarity of triangles
Since
AB ∥ CD,
triangles AEB and DEC are similar.
Hence,
AB/CD = AE/DE = BE/CE
Substituting the given values,
AE/DE = BE/CE = 2/1
Therefore,
AE = 2DE
and
BE = 2CE
Step 2: Express AD and BC
Since D lies on AE,
AD = AE − DE
= 2DE − DE
= DE
Thus,
AD = DE.
Similarly,
BC = BE − CE
= 2CE − CE
= CE
Thus,
BC = CE.
Hence,
AE = 2AD
and
BE = 2BC.
Step 3: Use the perimeter of the trapezium
Perimeter of ABCD
= AB + BC + CD + AD
6 = 2 + BC + 1 + AD
AD + BC = 3
Step 4: Find the perimeter of △AEB
Perimeter
= AE + BE + AB
= 2AD + 2BC + 2
= 2(AD + BC) + 2
= 2 × 3 + 2
= 8 cm
Answer:
A. 8
ABCD is a trapezium in which AB is parallel to DC, AD is perpendicular to AB, and AB = 3DC. If a circle inscribed in the trapezium touching all the sides has a radius of 3 cm, then the area, in sq. cm, of the trapezium is
48
30√3
36√2
54
48
Step 1: Let DC = x, AB = 3x.
AD ⊥ AB and incircle radius = 3 cm → AD = 6 cm (= diameter).
Step 2: BC = √((2x)² + 6²) = √(4x² + 36)
Step 3: Tangential quadrilateral property: AB + DC = AD + BC
4x = 6 + √(4x² + 36)
√(4x² + 36) = 4x − 6
Step 4: Square both sides:
4x² + 36 = 16x² − 48x + 36
12x² − 48x = 0 → x = 4
DC = 4 cm, AB = 12 cm
Step 5: Area = (1/2) × (12 + 4) × 6 = 48 sq. cm
A quadrilateral ABCD is inscribed in a circle such that AB : CD = 2 : 1 and BC : AD = 5 : 4. If AC and BD intersect at the point E, then AE : CE equals
2 : 1
1 : 2
8 : 5
5 : 8
8 : 5
For a cyclic quadrilateral, if the diagonals intersect at E, then
AE/CE = (AB × AD)/(CB × CD)
(This is a standard result obtained using similar triangles.)
Given, AB : CD = 2 : 1, BC : AD = 5 : 4
Let, AB = 2k, CD = k, BC = 5m, AD = 4m
Now, AE/CE = (AB × AD)/(BC × CD) = (2k × 4m)/(5m × k) = 8/5
Therefore, AE : CE = 8 : 5
Answer:
C. 8 : 5
In a rectangle ABCD, AB = 9 cm and BC = 6 cm. P and Q are two points on BC such that the areas of the figures ABP, APQ, and AQCD are in geometric progression. If the area of the figure AQCD is four times the area of triangle ABP, then BP : PQ : QC is
1:2:4
1:2:1
2:4:1
1:1:2
2:4:1
Step 1: Total area of rectangle
AB = 9 cm, BC = 6 cm → Area = 9 × 6 = 54 sq cm
Step 2: Determine areas using geometric progression
Let A1 = Area(ABP), A2 = Area(APQ), A3 = Area(AQCD).
A1, A2, A3 are in GP with common ratio r, and A3 = 4 × A1.
A1 × r² = 4 × A1 → r² = 4 → r = 2
So: A1 = a, A2 = 2a, A3 = 4a
Step 3: Solve for a
a + 2a + 4a = 54 → 7a = 54 → a = 54/7 sq cm
Step 4: Find segment lengths
Triangles ABP and APQ have height AB = 9 cm.
BP: Area(ABP) = (1/2) × BP × 9 → 54/7 = 4.5 × BP → BP = 12/7 cm
PQ: Area(APQ) = (1/2) × PQ × 9 → 108/7 = 4.5 × PQ → PQ = 24/7 cm
QC: BC − BP − PQ = 6 − 12/7 − 24/7 = 42/7 − 36/7 = 6/7 cm
Step 5: Ratio
BP : PQ : QC = 12/7 : 24/7 : 6/7 = 12 : 24 : 6 = 2 : 4 : 1
If the length of a side of a rhombus is 36 cm and the area of the rhombus is 396 sq. cm, then the absolute value of the difference between the lengths, in cm, of the diagonals of the rhombus is
Let diagonals be d₁ and d₂. Area = (1/2)d₁d₂ = 396 → d₁d₂ = 792.
Side relation: (d₁/2)² + (d₂/2)² = 36² → d₁² + d₂² = 5184.
(d₁−d₂)² = d₁² + d₂² − 2d₁d₂ = 5184 − 1584 = 3600. |d₁−d₂| = 60.
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