Question

1. what is the length of segment ab?

options:

• $10\sqrt{3}$cm
• $20\sqrt{3}$cm
• 10cm
• 20cm

Explanation:

Step1: Analyze Triangle XYC

In right triangle \( XYC \), \( \angle XYC = 90^\circ \), \( \angle XCY = 45^\circ \), so it's an isosceles right triangle. Thus, \( XY = YC = 5\,\text{cm} \)? Wait, no, wait. Wait, \( YC \) is given? Wait, no, the length \( BC = 10\,\text{cm} \), and \( XY \) is parallel to \( BC \)? Wait, \( XY \perp AC \) and \( BC \perp AC \), so \( XY \parallel BC \). So triangle \( AXY \sim \) triangle \( ABC \) (by AA similarity, since \( \angle A \) is common and \( \angle AYX = \angle ACB = 90^\circ \)).

First, look at triangle \( XYC \): \( \angle XYC = 90^\circ \), \( \angle XCY = 45^\circ \), so \( XY = YC \). Wait, but \( YC \) is part of \( AC \). Wait, \( BC = 10\,\text{cm} \), and \( XY \) is parallel to \( BC \), so \( XY = 10\,\text{cm} \)? Wait, no, let's re-examine.

Wait, \( XY \) is a horizontal segment, \( Y \) is on \( AC \), \( X \) is connected to \( B \) and \( C \). Wait, \( \angle XCY = 45^\circ \), \( \angle XYC = 90^\circ \), so \( \triangle XYC \) is isosceles right-angled, so \( XY = YC \). But \( BC = 10\,\text{cm} \), and \( XY \parallel BC \), so \( XY = BC = 10\,\text{cm} \)? Wait, no, \( BC \) is 10 cm, \( XY \) is parallel to \( BC \), so \( XY = 10\,\text{cm} \). Then in triangle \( AXY \), \( \angle A = 30^\circ \), \( \angle AYX = 90^\circ \), so \( \sin 30^\circ = \frac{XY}{AB} \)? Wait, no, \( \sin 30^\circ = \frac{XY}{AX} \)? Wait, no, let's clarify.

Wait, \( \angle A = 30^\circ \), \( \angle AYX = 90^\circ \), so in \( \triangle AYX \), \( \sin 30^\circ = \frac{XY}{AX} \)? No, \( \sin 30^\circ = \frac{opposite}{hypotenuse} = \frac{XY}{AX} \)? Wait, no, \( \angle A = 30^\circ \), so \( \sin 30^\circ = \frac{XY}{AX} \)? Wait, no, \( \angle A = 30^\circ \), \( \angle AYX = 90^\circ \), so \( \sin 30^\circ = \frac{XY}{AX} \)? Wait, no, \( XY \) is the opposite side to \( \angle A \)? Wait, no, \( \angle A = 30^\circ \), so the side opposite \( 30^\circ \) in \( \triangle AYX \) is \( XY \), and the hypotenuse is \( AX \). But we need to find \( AB \).

Wait, alternatively, in \( \triangle ABC \), \( \angle A = 30^\circ \), \( \angle C = 90^\circ \), so \( BC = \frac{1}{2}AB \) (since in a 30-60-90 triangle, the side opposite 30° is half the hypotenuse). Wait, \( BC = 10\,\text{cm} \)? Wait, no, \( BC \) is 10 cm? Wait, the diagram shows \( BC = 10\,\text{cm} \), and \( \angle A = 30^\circ \), \( \angle C = 90^\circ \), so \( BC \) is opposite \( \angle A = 30^\circ \), so \( BC = \frac{1}{2}AB \). Therefore, \( AB = 2 \times BC = 2 \times 10 = 20\,\text{cm} \). Wait, that makes sense. Let's check:

In right triangle \( ABC \), \( \angle A = 30^\circ \), \( \angle C = 90^\circ \), so the side opposite \( 30^\circ \) is \( BC \), so \( BC = \frac{1}{2}AB \). Given \( BC = 10\,\text{cm} \), then \( AB = 2 \times 10 = 20\,\text{cm} \).

Step2: Confirm with Similar Triangles

Since \( XY \parallel BC \), \( \triangle AXY \sim \triangle ABC \). \( \angle A = 30^\circ \), so \( \angle AXY = \angle ABC = 60^\circ \). In \( \triangle XYC \), \( \angle XCY = 45^\circ \), but maybe that's a red herring. The key is the 30-60-90 triangle ratio: in a right triangle, if one angle is 30°, the sides are in the ratio \( 1 : \sqrt{3} : 2 \) (opposite 30°, 60°, hypotenuse). Here, \( BC \) is opposite 30°, so \( BC = \frac{1}{2}AB \), so \( AB = 2 \times BC = 2 \times 10 = 20\,\text{cm} \).

Answer:

\( 20\,\text{cm} \) (the option is "20cm")