for exercises 24–27, find each length and angle measure for rectangle g h j k. round to the nearest tenth. see examples 3 and 4
24. ( mangle ghk )
25. ( mangle hlj )
26. ( gj )
27. ( hl )
(image of rectangle g h j k with h j = 7, j k = 10, ( angle k g j = 52^circ ), and diagonals h k and g j intersecting at l)

Question

Accepted Answer

### 24. $ m\angle GHK $
# Explanation:
## Step 1: Recall rectangle properties
In a rectangle, all angles are $ 90^\circ $, and the diagonals are equal and bisect each other. Also, opposite sides are equal. In rectangle $ GHJK $, $ \angle GKJ = 90^\circ $, and we know $ \angle KGH = 52^\circ $. In triangle $ GHK $, $ \angle GHK + \angle KGH + \angle GKH = 180^\circ $, but since $ \angle GKH = 90^\circ $ (rectangle angle), we can find $ \angle GHK $ as $ 90^\circ - 52^\circ $.
$$
m\angle GHK = 90^\circ - 52^\circ = 38^\circ
$$
# Answer:
$ 38.0^\circ $ (rounded to nearest tenth, though it's a whole number here)

### 25. $ m\angle HLJ $
# Explanation:
## Step 1: Diagonals in rectangle
In a rectangle, diagonals are equal and bisect each other, so $ GL = KL = HL = JL $. Also, triangle $ GHK $ and $ JHK $ are congruent. We know $ \angle GHK = 38^\circ $ (from problem 24) and $ \angle HJK = 52^\circ $ (since $ \angle KGH = 52^\circ $ and alternate interior angles or rectangle properties). The diagonals intersect at $ L $, so $ \triangle HLJ $ has $ HL = JL $, making it isosceles? Wait, no, actually, the angle at $ L $: first, find the angle between the diagonals. The diagonals in a rectangle, when we look at the angles, the sum of angles around $ L $ is $ 360^\circ $, but also, in triangle $ GKL $, $ \angle GLK = 180^\circ - 52^\circ - 52^\circ = 76^\circ $, and $ \angle HLJ $ is vertical to $ \angle GLK $? Wait, no, actually, $ \angle HLJ $ is supplementary to the angle we can find. Wait, another approach: the diagonals intersect, and the angle $ \angle HLJ $ is equal to $ 180^\circ - 2\times38^\circ $? No, wait, let's use the fact that in rectangle, the triangles formed by diagonals are congruent. Alternatively, since $ \angle KGH = 52^\circ $, and $ GH $ is parallel to $ JK $, the angle between diagonal $ HJ $ and $ GH $ is $ 38^\circ $, but when diagonals intersect at $ L $, $ \angle HLJ $ is the angle between the two diagonals. Wait, actually, the correct way: in triangle $ GHK $, we know $ \angle GHK = 38^\circ $, $ \angle HGK = 52^\circ $. The diagonals bisect each other, so $ GL = HL $, so triangle $ GLH $ is isosceles with $ GL = HL $, so $ \angle LGH = \angle LHG = 52^\circ $? No, wait, no. Wait, $ \angle KGH = 52^\circ $, which is $ \angle LGH = 52^\circ $, and $ GL = HL $, so $ \angle LHG = \angle LGH = 52^\circ $, so $ \angle GLH = 180 - 52 - 52 = 76^\circ $, then $ \angle HLJ = 180 - 76 = 104^\circ $? Wait, no, that's not right. Wait, actually, the diagonals in a rectangle: the angle between the diagonals can be found by considering the triangles. Wait, maybe a better way: the sum of angles in $ \triangle HLJ $. Wait, no, let's use the fact that $ \angle HLJ $ is equal to $ 180^\circ - 2\times38^\circ $? No, wait, let's calculate the length of the diagonal first (maybe from problem 26) and use trigonometry, but maybe easier: in rectangle, the angle $ \angle HLJ $ is $ 180^\circ - 2\times(90^\circ - 52^\circ) $? Wait, no, let's go back. The angle at $ G $ is $ 52^\circ $, so the angle between $ GH $ and $ GK $ is $ 90^\circ $, so $ \angle HGK = 52^\circ $, so $ \angle GHK = 38^\circ $. The diagonals intersect at $ L $, so $ HL = GL $, so triangle $ GLH $ has $ \angle GLH = 180^\circ - 52^\circ - 52^\circ = 76^\circ $, and $ \angle HLJ $ is supplementary to $ \angle GLH $? No, $ \angle HLJ $ and $ \angle GLK $ are vertical angles, and $ \angle GLK = 76^\circ $? Wait, no, I think I made a mistake. Let's use the fact that in the rectangle, the diagonals are equal, so $ GJ = HK $, and we can find the length of the diagonal (problem 26) as $ \sqrt{7^2 + 10^2} = \sqrt{49 + 100} = \sqrt{149} \approx 12.2 $. Then, in triangle $ HLJ $, $ HL = JL = \frac{12.2}{2} \approx 6.1 $, and $ HJ = 7 $? Wait, no, $ HJ $ is 7? Wait, the diagram: $ HJ = 7 $, $ JK = 10 $. So $ GH = JK = 10 $, $ HJ = GK = 7 $. Wait, no, rectangle: $ GH $ and $ JK $ are length 10, $ HJ $ and $ GK $ are length 7. So diagonals $ GJ $ and $ HK $ are equal, length $ \sqrt{7^2 + 10^2} = \sqrt{149} \approx 12.20655 $. Then, in triangle $ GHK $, which is a right triangle with legs 7 and 10, angle at $ G $ is $ 52^\circ $, so angle at $ H $ is $ 38^\circ $. Now, diagonals intersect at $ L $, so $ HL = \frac{HK}{2} \approx \frac{12.20655}{2} \approx 6.103 $. Now, in triangle $ HLJ $, we have $ HL = JL \approx 6.103 $, and $ HJ = 7 $. Using the Law of Cosines: $ HJ^2 = HL^2 + JL^2 - 2 \times HL \times JL \times \cos(\angle HLJ) $. So $ 7^2 = 6.103^2 + 6.103^2 - 2 \times 6.103 \times 6.103 \times \cos(\angle HLJ) $. Calculate $ 49 = 2 \times 6.103^2 - 2 \times 6.103^2 \times \cos(\angle HLJ) $. $ 6.103^2 \approx 37.25 $, so $ 2 \times 37.25 = 74.5 $. So $ 49 = 74.5 - 74.5 \times \cos(\angle HLJ) $. Then, $ 74.5 \times \cos(\angle HLJ) = 74.5 - 49 = 25.5 $. So $ \cos(\angle HLJ) = \frac{25.5}{74.5} \approx 0.3423 $. Then $ \angle HLJ = \arccos(0.3423) \approx 70.0^\circ $? Wait, no, that can't be. Wait, I think I mixed up the sides. Wait, $ HJ $ is 7, $ JK $ is 10, so $ GH = 10 $, $ HJ = 7 $. So diagonals are $ \sqrt{10^2 + 7^2} = \sqrt{149} \approx 12.20655 $. Then, in triangle $ GHK $, right-angled at $ H $? Wait, no, rectangle: $ \angle H = 90^\circ $, so $ \triangle GHK $ is right-angled at $ H $, with $ GH = 10 $, $ HK = 7 $? Wait, no, the diagram: $ HJ = 7 $, $ JK = 10 $, so $ HJ $ is horizontal, $ JK $ is vertical. So $ H $ to $ J $ is 7 (horizontal), $ J $ to $ K $ is 10 (vertical), so $ H $ to $ K $ is diagonal, length $ \sqrt{7^2 + 10^2} \approx 12.2 $. $ G $ to $ K $ is 7 (horizontal), $ G $ to $ H $ is 10 (vertical). So diagonals $ GJ $ and $ HK $ are both $ \sqrt{7^2 + 10^2} \approx 12.2 $. Now, angle $ \angle GHK $: in $ \triangle GHK $, right-angled at $ H $, so $ \tan(\angle GHK) = \frac{GK}{GH} = \frac{7}{10} = 0.7 $, so $ \angle GHK = \arctan(0.7) \approx 35^\circ $? Wait, no, earlier I thought $ \angle KGH = 52^\circ $, which is given in the diagram: $ \angle KGH = 52^\circ $. So $ \angle KGH = 52^\circ $, so in $ \triangle GHK $, right-angled at $ H $, $ \angle KGH = 52^\circ $, so $ \angle GHK = 90^\circ - 52^\circ = 38^\circ $, which matches. Now, diagonals intersect at $ L $, so $ GL = LJ = HL = LK $. So $ \triangle GLH $ has $ GL = HL $, so it's isosceles with $ \angle LGH = \angle LHG = 52^\circ $, so $ \angle GLH = 180^\circ - 52^\circ - 52^\circ = 76^\circ $. Then $ \angle HLJ $ is supplementary to $ \angle GLH $? No, $ \angle HLJ $ and $ \angle GLK $ are vertical angles, and $ \angle GLK = 76^\circ $? Wait, no, $ \angle HLJ $ is actually equal to $ 180^\circ - 76^\circ = 104^\circ $? Wait, let's use the fact that the sum of angles around point $ L $ is $ 360^\circ $, and opposite angles are equal. So $ \angle GLH = \angle KLJ $, $ \angle HLK = \angle GLJ $. We found $ \angle GLH = 76^\circ $, so $ \angle HLK = 180^\circ - 76^\circ = 104^\circ $? No, that's not right. Wait, I think the correct answer is $ 104^\circ $, because $ \angle HLJ $ is the angle between the two diagonals, and since $ \angle GHK = 38^\circ $, and $ HL = JL $, the triangle $ HLJ $ has angles: $ \angle LHJ = \angle LJH = 38^\circ $, so $ \angle HLJ = 180^\circ - 38^\circ - 38^\circ = 104^\circ $. Yes, that makes sense. So $ m\angle HLJ = 104.0^\circ $ (rounded to nearest tenth).
## Step 1: Angles in isosceles triangle
Since $ HL = JL $ (diagonals bisect each other), $ \triangle HLJ $ is isosceles with $ \angle LHJ = \angle LJH = 38^\circ $ (from $ \angle GHK = 38^\circ $).
$$
m\angle HLJ = 180^\circ - 38^\circ - 38^\circ = 104^\circ
$$
# Answer:
$ 104.0^\circ $

### 26. $ GJ $
# Explanation:
## Step 1: Pythagorean theorem in rectangle
In a rectangle, the diagonal can be found using the Pythagorean theorem, where the diagonal is the hypotenuse of a right triangle with legs equal to the length and width of the rectangle. Here, the length is $ 10 $ ( $ GH = JK = 10 $ ) and the width is $ 7 $ ( $ HJ = GK = 7 $ ). So, for diagonal $ GJ $, we use $ GJ = \sqrt{GH^2 + HJ^2} $.
$$
GJ = \sqrt{10^2 + 7^2} = \sqrt{100 + 49} = \sqrt{149} \approx 12.20655 \approx 12.2
$$
# Answer:
$ 12.2 $

### 27. $ HL $
# Explanation:
## Step 1: Diagonals in rectangle
In a rectangle, diagonals are equal and bisect each other, so $ HL = \frac{HK}{2} $. We already found $ HK = \sqrt{7^2 + 10^2} = \sqrt{149} \approx 12.20655 $ (from problem 26, since $ HK = GJ $ in a rectangle). So, $ HL = \frac{\sqrt{149}}{2} \approx \frac{12.20655}{2} \approx 6.103275 \approx 6.1 $.
$$
HL = \frac{\sqrt{149}}{2} \approx \frac{12.2066}{2} \approx 6.1
$$
# Answer:
$ 6.1 $

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QUESTION IMAGE

Question

Explanation:

24. \( m\angle GHK \)

Step 1: Recall rectangle properties

Step 1: Diagonals in rectangle

Step 1: Pythagorean theorem in rectangle

Answer:

25. \( m\angle HLJ \)