Question: Chapter 10.2
Pusat Tuisyen Gaya Pintas
1.
Dalam rajah di bawah, KLMN ialah sebuah trapezium dan KLOP ialah sebuah segi empat selari.
In the diagram, KLMN is a trapezium and KLOP is a parallelogram.
K L = 10 cm
K P = 6 cm
N M = 16 cm
12 cm
Cari luas, dalam cm², kawasan yang berlorek.
Find the area, in cm², of the shaded region.
A 900
B 146
C 96
D 84
2.
(a) Rajah di bawah menunjukkan sebuah trapezium PQRS. Luas segi tiga PQT ialah 30 cm².
The diagram shows a trapezium PQRS. The area of triangle PQT is 30 cm².
P Q = x cm
P T = 13 cm
P S = 10 cm
S R = 18 cm
Q T = 5 cm
T R = 3 cm
(i) Cari nilai x.
Find the value of x.
[2 markah/2 marks]
Jawapan/Answer:
(ii) Hitung luas, dalam cm², kawasan berlorek.
Find the area, in cm², of the shaded region.
[2 markah/2 marks]
Jawapan/Answer:
3.
Rajah di bawah menunjukkan sebuah trapezium KLMN.
The diagram shows a trapezium KLMN.
K L = 10 cm
K N = 6 cm
N M = 12 cm
Hitung luas, dalam cm², kawasan berlorek.
Find the area, in cm², of the shaded region.
[3 markah/3 marks]
Jawapan/Answer:
Graph shape: a trapezium KLMN in the bottom-left with shaded interior and two line segments from a point on KN to points on NM; position_hint=bottom-left.
Graph shape: a trapezium PQRS in the top-right with a triangular shaded region and a segment PQ labeled x cm, with QT = 5 cm and TR = 3 cm; position_hint=top-right.User: Help me answer thank you
1. **Problem:** Find the area of the shaded region in trapezium KLMN where KLOP is a parallelogram.
Given:
- $KL = 10$ cm
- $KP = 6$ cm
- $NM = 16$ cm
- Height (distance between parallel sides) = $12$ cm
**Step 1:** Calculate the area of trapezium KLMN.
Formula for trapezium area:
$$\text{Area} = \frac{1}{2} \times (\text{sum of parallel sides}) \times \text{height}$$
Here, parallel sides are $KL = 10$ cm and $NM = 16$ cm.
$$\text{Area}_{KLMN} = \frac{1}{2} \times (10 + 16) \times 12 = \frac{1}{2} \times 26 \times 12 = 13 \times 12 = 156 \text{ cm}^2$$
**Step 2:** Calculate the area of parallelogram KLOP.
Base $KP = 6$ cm and height is the same $12$ cm.
$$\text{Area}_{KLOP} = \text{base} \times \text{height} = 6 \times 12 = 72 \text{ cm}^2$$
**Step 3:** The shaded area is the difference between trapezium and parallelogram areas.
$$\text{Shaded area} = 156 - 72 = 84 \text{ cm}^2$$
**Answer:** $84$ cm² (Option D)
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2. (a) **Problem:** Given trapezium PQRS with triangle PQT area $30$ cm², find $x = PQ$ and the shaded area.
Given:
- Area of $\triangle PQT = 30$ cm²
- $PT = 13$ cm
- $PS = 10$ cm
- $SR = 18$ cm
- $QT = 5$ cm
- $TR = 3$ cm
(i) Find $x = PQ$.
**Step 1:** Area of triangle formula:
$$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$$
Here, base $PT = 13$ cm, height is $PQ = x$ cm.
$$30 = \frac{1}{2} \times 13 \times x$$
Multiply both sides by 2:
$$60 = 13x$$
Divide both sides by 13:
$$x = \frac{60}{13} \approx 4.615$$
(ii) Find the shaded area.
**Step 2:** The shaded region is trapezium PQRS minus triangle PQT.
Calculate trapezium area:
Parallel sides: $PQ = x = \frac{60}{13}$ cm and $SR = 18$ cm.
Height is $PT + TR = 13 + 3 = 16$ cm.
$$\text{Area}_{PQRS} = \frac{1}{2} \times (x + 18) \times 16 = 8 \times \left(\frac{60}{13} + 18\right) = 8 \times \frac{60 + 234}{13} = 8 \times \frac{294}{13} = \frac{2352}{13} \approx 180.92$$
**Step 3:** Shaded area = trapezium area - triangle area
$$180.92 - 30 = 150.92 \text{ cm}^2$$
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3. **Problem:** Find the area of the shaded region in trapezium KLMN.
Given:
- $KL = 10$ cm
- $KN = 6$ cm
- $NM = 12$ cm
**Step 1:** Calculate the height of trapezium using Pythagoras or given data (assuming height is $6$ cm as $KN$ is height).
**Step 2:** Calculate area of trapezium:
$$\text{Area} = \frac{1}{2} \times (KL + NM) \times KN = \frac{1}{2} \times (10 + 12) \times 6 = \frac{1}{2} \times 22 \times 6 = 11 \times 6 = 66 \text{ cm}^2$$
**Answer:** $66$ cm²